Given that $W$ is a subspace of $V$, prove: $\dim W=\dim V \implies W=V$ Suppose $W$ is a subspace of the finite-dimensional vector space $V$ and $\dim W = \dim V=n$
Let $\{\overrightarrow w_1,\ldots,\overrightarrow w_n\}$ be a basis for $W$ and $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ be a basis for $V$.
The set $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ still spans $W$. Since $W$ is a subspace of $V$, $W\subset V$ and all the $w_i$'s can be written in terms of $\{ \overrightarrow v_{1},\ldots,\overrightarrow v_n\}$.
$\therefore$ We can remove one by one all the $w_i$'s from the set $\{\overrightarrow w_{1},\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ while still having it span $W$, which leaves us with the set $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$. We know it to be linearly independent since it is a basis for $V$. Therefore the basis of $V$ is also a basis of $W$. From there on, it is trivial to prove double inclusion $W\subset V$ and $V\subset W$ which gives us $W=V$.
Is my approach correct?
 A: Your approach is incorrect: the set $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ spans $V$ and saying that it spans $W$ is the same as assuming $W=V$.
You should rather consider a basis $\{\overrightarrow w_1,\ldots,\overrightarrow w_n\}$ of $W$ and assume, by contradiction, that $W\ne V$. Then there exists $\vec{v}\in V$ with $\vec{v}\notin W$. It's easy to prove that $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v\}$ is linearly independent and now this is a contradiction, because we found a linearly independent set in $V$ with more elements than $\dim V$.
A: Your approach is fine and correct. 
The easiest way to prove this is to notice that $W$ is a subspace of $V$, thus it is a subset of $V$. Thus, if $dimV = n$, and we have $n$ linearly independent vectors in $W$ that span $W$, these must also be linearly independent vectors in $V$. But then we have $n$ of them so these must span $V$ as well! So the vectors span $W$ AND $V$, so they are equal.
A: Your proof is incorrect. You first choose a colloquial understanding of the word "spanning" and at a later point the mathematically correct understanding [which changes the meaning of the word!].
You say that $\{w_1, \ldots, w_n, v_1, \ldots, v_n\}$ spans $W$, but you can't deduce this directly. What you mean is that $W$ is a subset of $\operatorname{span}(\{w_1, \ldots, w_n, v_1, \ldots, v_n\})$. Since the $w_i$ can be represented as linear combinations of the $v_i$, you can succesively remove the $w_i$ without changing the span, i.e.
$$\operatorname{span}(\{w_1, \ldots, w_n, v_1, \ldots, v_n\}) = \operatorname{span}(\{v_1, \ldots, v_n\})$$
Now at this point in your argumentation, you make the switch from the colloquial term "spanning" to the mathematically rigorous by saying that $\{v_1, \ldots, v_n\}$ spans $W$ and is therefore a basis for $W$. But this is incorrect, as you've only shown that $\{v_1, \ldots, v_n\}$ "overspans" $W$, i.e. $W \subset \operatorname{span}(\{v_1, \ldots, v_n\})$. But this inclusion is trivial, since the span of the $v_i$ is the whole of $V$.
Note that you've at no point in your argumentation actually used the fact that the dimensions of $V$ and $W$ are the same. Simply go throgh the proof and try to point at the part where you've used this assumption.
