If $V \subseteq W$ and dim$(V)$=dim$(W)$, then $V=W$ If $V,W$ are subspaces of $\mathbb{R^n}$.
My proof simply states that if the dimension, $m$, of both subspaces are the same, then we know that $m$ linearly independent vectors in both $V$ and $W$ span $V$ and $W$. Thus, if $V \subseteq W$, these $m$ linearly independent vectors must be the same, and so the span of both subspaces must be the same $\Rightarrow$ $V = W$.
Is this the type of rigor expected in such a proof? I can see conceptually why the claim is true, but I'm not really sure how to organize my thoughts in such a way that "proves" the claim.
 A: You have some of the right ideas for a proof, but your wording is kind of confused and jumbled. A good proof should follow clearly from "known" facts/theorems.
Your idea about spanning is a good one. Start there. 
Take a basis, $\beta$, for $V$. This consists of $m = \dim(V)$ linearly independent vectors which span $V$. Then since $V \subseteq W$, you have that the vectors in $\beta$ belong to $W$. So $\beta$ is a collection of $n$ linearly independent vectors in $W$. Thus (since $\dim(W)=m$) this collection is a basis and thus spans $W$. Now you have
$$V = \mathrm{span}(\beta) = W$$
done. :)
A: Extend a basis $v_1,\dotsc,v_m$ of $V$ to a basis $v_1,\dotsc,v_{m+n}$ of $W$. Then 
$$
m=\dim V=\dim W=m+n
$$
so $n=0$. It follows that $v_1,\dotsc,v_m$ is also a basis for $W$. Hence $V=W$.
A: Your argument is correct. However, to make things completely clear you can put it in this way. If $W$ is a proper subspace of $V$, then we can choose a basis for $W$ and extend it to a basis of $V$ by adding vectors to it. But this is impossible since $\dim V=\dim W$.
