TITLE: Proof about inclusion of subspaces
QUESTION [5 upvotes]: I have the following question and I'm not sure my proof is correct/approached correctly:

Let $V$ be a $n$-dimensional vector space, let $U_i \subset V$ be subspaces of V for $i = 1,2,\dots,r$ where $$U_1 \subset U_2 \subset \dots \subset U_r$$
  If $r>n+1$ then there exists an $i<r$ for which $U_i = U_{i+1}$

I was thinking something along the lines of: the subsets are strict thus 'moving' from $U_i$ to $U_{i+1}$ increases the dimension by one, which means, when reaching $U_r$, your dimension is greater than $n+1$ but this is not possible since $U_r$ is a subspace of $V$ and thus has dimension at most $n$.
Is this a good approach to this proof? I'm not sure because of the strict inclusions: is it true for every case that $dim(U_i) < dim(U_{i+1})$?
I was maybe thinking of proving this via induction, but I'm not sure how that would work.
Thanks in advance.
EDIT: Updated subset-strictness

REPLY [3 votes]: Your idea is very good, although not well expressed; but also the statement is not well written. Let's state it properly.

Let $V$ be a $n$-dimensional vector space, let $U_i \subseteq V$ be subspaces of $V$ for $i = 1,2,\dots,r$ where $$U_1 \subseteq U_2 \subseteq \dots \subseteq U_r$$
  If $r>n+1$ then there exists an $i<r$ for which $U_i = U_{i+1}=\dots=U_r$

Let's prove, by induction on $r$, that, given a sequence of subspaces
$$
U_1\subsetneq U_2\subsetneq\dots\subsetneq U_r
$$
then $\dim U_r\ge r-1$. We only use the fact that $U\subsetneq U'$ implies $\dim U'\ge 1+\dim U$.
If $r=1$, this is trivial. Assume it holds for $r-1$. Then, by induction hypothesis, $\dim U_{r-1}\ge r-2$ and so $\dim U_r\ge (r-2)+1=r-1$.
So, if we have
$$
U_1\subseteq U_2\subseteq\dots\subseteq U_r
$$
and $r>n+1$, the above result proves that not all inclusions are proper. 
Let $j$ be the maximum index such that $U_{j}\subsetneq U_{j+1}$ (or $j=0$ if all inclusions are equalities): by the result above, $j<r-1$. Now, for $i=j+1$, $U_i=U_{i+1}=\dots=U_r$.