TITLE: Forcing as a replacement of induction and diagonal arguments
QUESTION [33 upvotes]: Let me give some examples motivating the question.
The use of forcing instead of induction: For this consider Cantor's theorem:
Theorem 1. Any two countable dense linear orders $I, J$ without end points are order isomorphic. 
Proof. Let $\mathbb{P}$ be the partial order consisting of partial finite order preserving maps from $I$ to $J$. For $i\in I, j\in J$ the sets $D_i=\{ p: i\in dom(p)\}$ and $R_j=\{p: j\in range(p)   \}$ are dense, and since we have only countably many dense sets, we can get a filter $G$ which meets all of these dense sets. Then $\bigcup G: I \to J$ is the required isomorphism.
The use of forcing instead of diagonal arguments: This time let's consider another result of Cantor:
Theorem 2. For any regular cardinal $\kappa, 2^\kappa > \kappa.$
Proof. Let $F$ be a family of functions from $\kappa\to 2$ of size $\kappa.$ Let $Add(\kappa, 1)$ be the Cohen forcing for adding a new subset of $\kappa,$ and for $f\in F,$ let $D_f=\{p: p\neq f   \}.$ Each $D_f$ is easily seen to be dense in $Add(\kappa, 1)$, and since the forcing is $\kappa-$closed, there is an $F-$generic filter $G$ over $Add(\kappa, 1).$ Then $\bigcup G: \kappa \to 2$ is different from all $f\in G.$
Theorem 3. If $\lambda=cf(\kappa) < \kappa,$ then $\kappa^\lambda > \kappa.$
Proof. Let $F$ be family of size $\kappa$ of functions from $\lambda\to \kappa.$ Let $\mathbb{P}=\{p:p$ is a partial function of size $<\lambda$ from $\lambda\to \kappa   \}.$ It is $\lambda-$closed. Let $(\kappa_\alpha: \alpha<\lambda)$ be increasing cofinal in $\kappa,$ let $F_0 \subseteq F_1 \subseteq ... (\alpha<\lambda)$ with $|F_\alpha|=\kappa_\alpha$ and $F=\bigcup_\alpha F_\alpha.$ Now consider dense sets $D_\alpha=\{p: \forall f\in F_\alpha, p\neq f  \}.$ If a filter $G$ meets all $D_\alpha$'s, then $\bigcup G$ is different from all elements of $F$.
Below are a few more examples of the results that can be proved in a similar way:
(A) If $\mathcal{A}$ and $\mathcal{B}$ are countable families of subsets of
$\mathbb{N}$ such that $A \cap B$ is finite for each $A\in \mathcal{A}$ and $B\in \mathcal{B}$, then there is
a $C$ such that $A \subseteq^*  C$ for every $A\in \mathcal{A}$ and $B\cap C$ is finite for every $B\in \mathcal{B}.$
(B) There exists  a continuous, nowhere dieffrentiable
function on $[0,1].$
(C)  Let's call a set $A$ of natural numbers (not including 0) is small, if $\Sigma_{n\in A}1/n < \infty.$ Given a countable family $\mathcal{F}$ of small sets, there exists a small set $J$ such that $I\subseteq^* J,$ for all $I\in \mathcal{F}.$
(ِِD) All consequences of $MA(\aleph_0),$ in particular the Baire category theorem,....

Question. Are there any more non-trivial results which are proved using forcing in the following way: we produce a forcing notion $\mathbb{P}$ and a family $\mathcal{D}$ of dense subsets of it. Then we argue that there must be a $\mathcal{D}-$generic filter $G$ over $\mathbb{P},$ and use $G$ to conclude our required result.

REPLY [6 votes]: There is a forcing proof of Sierpinski's theorem that $\mathbb{Q}$ with its usual topology is the unique countable, metrizable space without isolated points. The argument is analogous to the proof of Cantor's theorem given in the OP. In both cases, one uses a forcing poset to mimic a back-and-forth construction. 
Theorem (Sierpinski). If $X$ and $Y$ are countable, metrizable spaces without isolated points, then $X$ is homeomorphic to $Y$. 
Proof. Fix metrics for $X$ and $Y$. Since these spaces are countable, most of their closed balls are also open. Hence we may fix a base of clopen sets $\langle O_i: i \in \omega \rangle$ for $X$ and similarly $\langle N_j: j \in \omega \rangle$ for $Y$. The idea is to define a poset whose conditions approximate some bijection $h: X \rightarrow Y$, and arranges that $h[O_i]$ will be open in $Y$ and $h^{-1}[N_j]$ will be open in $X$ for every $i,j$. 
Let $\mathbb{P}$ be the poset with conditions of the form $(f, m, P, Q)$, where 

$P$ is a finite partition of $X$ into clopen sets,
$Q$ is a finite partition of $Y$ into clopen sets with $|Q| = |P|$,
$m$ is a bijection of $P$ with $Q$,
$f$ is a finite partial injection from $X$ into $Y$ that respects the matching $m$ (i.e. if $x \in$ dom($f$) and $p$ is the unique piece of the partition $P$ containing $x$, then $f(x)$ is in $m(p)$).

A condition $(f', m', P', Q')$ extends $(f, m, P, Q)$ if $P'$ refines $P$, $Q'$ refines $Q$, $m'$ refines $m$ (in the obvious sense), and $f'$ extends $f$. 
I claim that for every $x \in X$, $y \in Y$, and $i, j \in \omega$, the set of conditions $D_{x, y, i, j} = \{(f, m, P, Q)$: $x \in$ dom($f$), $y \in$ ran($f$), $O_i$ is a union of pieces in $P$, $N_j$ is a union of pieces in $Q\}$, is dense in $\mathbb{P}$. To see this, fix a condition $(f, m, P, Q)$. Given $x \not \in$ dom($f$), there is a unique piece $p_x$ in $P$ containing $x$. Since $Y$ is without isolated points, $m(p_x)$ is infinite. Pick $z \in m(p_x)$ not already in ran($f$). Similarly, given $y \not \in$ ran($f$), we may find $w \in m^{-1}(q_y)$ not already in the dom($f$), where $q_y$ is the unique piece of $Q$ containing $y$. Then $f' = f \cup \{(x, z), (w, z)\}$ extends $f$ and still respects $m$. 
Now, any fixed $O_i$ splits every partition piece $p$ into two clopen sets, $p_1 = O_i \cap p$ and $p_2 = (X \setminus O_i) \cap p$. Let $P'$ be the partition consisting of all sets of this form (ignoring those intersections which are empty). Then $P'$ refines $P$ and $O_i$ is a union of elements in $P'$. We must find a matching refinement of $Q$. Given $q \in Q$ we have $q = m(p)$ for some $p \in P$. We split $q$ into two pieces $q_1$ and $q_2$ as follows. If  $p_1 = p$ and $p_2 = \emptyset$, let $q_1 = q$ and $q_2 = \emptyset$. And vice versa. Now suppose both $p_1$ and $p_2$ are nonempty. If dom($f'$) $\cap \, p_1$ is empty, then let $q_1$ be a clopen ball strictly contained in $q$ that does not intersect ran($f'$). Let $q_2 = q \setminus q_1$. If instead dom($f'$) $\cap \, p_1$ is nonempty, for every $x \in$ dom($f'$) $\cap \, p_1$ pick a clopen ball around $f(x)$ that is strictly contained in $q$ and contains no other points in ran($f'$). Let $q_1$ be the union of these balls, and $q_2 = q \setminus q_1$. Let $Q'$ be the collection of the $q_i$ (again ignoring any empty sets), and $m'$ be the function that sends $p_i$ to $q_i$ for every $p$, $i$. Then $Q'$ refines $Q$ and $m'$ (which refines $m$) is a matching of $P'$ with $Q'$ that respects $f'$. 
Similarly, given any $N_j$ we may refine $P'$ to $P''$, $Q'$ to $Q''$ and $m'$ to $m''$ so that $N_j$ is a union of elements in $Q''$ and $m''$ still respects $f'$. Then $(f', m'', P'', Q'')$ is in $D_{x, y, i, j}$ and extends $(f, m, P, Q)$, as desired.  
Since there are only countably many $D_{x, y, i, j}$, we can find a filter $G$ meeting all of them. Taking the union over the $f$'s in $G$ yields a bijection $h: X \rightarrow Y$. This $h$ is actually a homeomorphism. For if $O_i$ is a fixed element of our base for $X$, there is a condition $(f, m, P, Q)$ in $G$ where $O_i$ is a union of pieces in $P$. These pieces are matched by $m$ to pieces in $Q$, and one may check that the union of these pieces (which is open, in fact clopen, in $Y$) is exactly $h[O_i]$. Similarly $h^{-1}[N_j]$ is open in $X$ for every $j$, so that $h$ is a homeomorphism, as claimed.
Cantor's theorem and Sierpinski's theorem have the following generalizations that are useful in certain contexts, and can be similarly proved by forcing arguments by modifying the respective posets in the evident way:
Theorem (Skolem): Fix some $k$, $1 \leq k \leq \omega$. Let $X, Y$ be countable dense, linear orders without endpoints. Fix a partition $X = \bigcup_{i < k} X_i$ such that each $X_i$ is dense in $X$, and similarly $Y=\bigcup_{i < k}Y_i$. There is an isomorphism $f: X \rightarrow Y$ such that $f[X_i]=Y_i$ for every $i < k$.
Theorem: Fix some $k$, $1 \leq k \leq \omega$. Let $X, Y$ be countable, metrizable spaces without isolated points. Fix a partition $X = \bigcup_{i < k} X_i$ such that each $X_i$ is dense in $X$, and similarly $Y=\bigcup_{i <k}Y_i$. There is a homeomorphism $f: X \rightarrow Y$ such that $f \upharpoonright X_i$ is a homeomorphism of $X_i$ with $Y_i$ for every $i<k$.