TITLE: $\Bbb Q$ is the unique countable and metrizable topological space without isolated points. But how?
QUESTION [5 upvotes]: This answer cites a result which states the following

Theorem. Let $X$ be a topological space which is

countable,
metrizable,
has no isolated points.

Then $X$ is homeomorphic to $\Bbb Q$ with the usual order topology.

But how is this even possible? I made no effort to understand the proof $-$ yet. Mostly because until now it seems more "obvious" that this is false. How about the following "counterexamples":

Example 1. $\Bbb Q$ and $\Bbb Q^2$ are both countable, metrizable, have no isolated points and are connected. But removing a single point diconnects $\Bbb Q$ but not $\Bbb Q^2$.
Example 2. What about $X=\Bbb Q^2\cap S^1$? It seems we can cover $X$ with connected open sets $U_i,i\in\Bbb Z_4$ so that $U_i\cap U_j=\varnothing$ if and only of $i-j=2$. Seems not possible for $\Bbb Q$.

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All my counterexamples use the term "connected". Are there some subtleties with this term which I am overlooking?

REPLY [2 votes]: It is my opinion that looking at metric spaces clouds the essence of how to characterize $\mathbb Q$ as a topological space. So we throw that out at the start of this probe...

The question/answer
Showing $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}^2$,
when seen for the first time has a bit of 'surprise element' to it; counter-intuitive or just weird, take your pick.
Applying the 'forgetful functor' to $\mathbb{Q}$, stripping away everything but the topology is a fun thing to think about. I've been on that journey the past few days; you might find this of interest:
Let $X$ be any countably infinite set with a well-ordering $X \equiv \mathbb N$,
$\quad s_0,s_1,s_2,s_3,\dots,s_n,\dots$
The element $s_0$ is in $X_0 = X$. We can certainly partition $X_0$ into two infinite disjoint sets $X_\text{00}$ (left) and $X_\text{01}$ (right) with $s_0 \in X_\text{00}$ and $s_1 \in X_\text{01}$. From this point on, we put the 'next element' to the 'left' or 'right' as we divide sets. So if $s_2 \in X_\text{00}$, divide $X_\text{00}$ into two disjoint infinite sets $X_\text{000}$ and $X_\text{001}$ with $s_2 \in X_\text{000}$ and $s_0 \in X_\text{001}$. 
So, when you finish processing $s_n$, the set $X$ will be partitioned into $n+1$ infinite sets each containing exactly one $s_k$ with $k \le n$. It will be a finer partition than the preceding recursion performed on $s_{n-1}$.
Proposition 1: The sets $X_\tau$ form a countable basis of sets, defining a topology on $X$.
Proof: Easy exercise.
I haven't worked out all the details, but once I checked and saw the $3^{rd} \text{-method}$ in the'3-proofs paper', this is worth some study:
Exercise: Show that $X$ is homeomorphic to $\mathbb Q$.
The set $X$ with this topology is, to my way of thinking, the purest point-set approach to describing the topology on $\mathbb Q$ - it is the 'canonical' set theoretic form. Knowing no theory, it will 'push you along' to discover interesting facts.