Let $V$ be a finite dimensional normed vector space, and $W$ be a subspace of $V$. Show that $\{\inf(\| v-w\|) : w \in W\}$ attains a minimum on $W$. 
Let $V$ be a finite dimensional normed vector space, and $W$ be a subspace of $V$. Show that $\{\inf(\| v-w\|) : w \in W\}$ attains a minimum on $W$.

So we want to show that $\exists w' \in W$ such that $\| v - w'\| = \inf\{\| v-w\| : w \in W\}$.
I'm pretty confused here. I have 2 potential ideas, but I can't seem to take them anywhere:
1) Define $ F:W \rightarrow \Bbb R $ by $ F(w) = \| v-w\| $. If we can show that $F$ is continuous on $W$, and if we can find a compact set $W'$ that contains $w'$, then we can apply EVT and we are done. I can't seem to show here that $F$ is continuous. And beyond that, I'm not sure how to assert the existence of such a compact set.
2) Show that every point in $W$ is an accumulation point. Then $\exists$ a sequence in $W$ that converges to $w'$. This sequence is then Cauchy in $W$, so that it  converges to $w'$. Then, since $W$ is complete, $w' \in W$, and the result follows by continuity of the norm. I am not sure if this approach is completely correct, or whether the initial assertion is correct at all.
Any hints are greatly appreciated!
 A: Here is one way to flesh out the details for your first idea. Let $F:W \to \mathbb{R}$ be as defined in the question.
Firstly, by the reverse triangle inequality, $|F(w) - F(w')| = \big|\|v-w\| - \|v-w'\| \big | \leq \|w-w'\|$ so $F$ is even Lipschitz continuous. 
So we are only left to find a suitable compact set. Since $V$ is finite dimensional, so is $W$ (and in fact, it is only $W$ that we need to be finite-dimensional here). Hence, by the Heine-Borel theorem, we know that the compact subsets of $W$ are exactly the closed and bounded sets. 
Pick an arbitrary $w_0 \in W$ and let $r = \|v-w_0\|$. Then we have that $$\inf \{F(w): w \in W\} = \inf\{F(w): w \in \overline{B}_V(v, r) \cap W\}$$
where $\overline{B}_V(v, r)$ is the closed ball in $V$ about $v$ of radius $r$. But $\overline{B}_V(v, r) \cap W$ is a closed and bounded set in $W$ and hence is compact. So since $F$ is continuous, by the extreme value theorem, there is $w' \in \overline{B}_V(v, r) \cap W$ such that $$F(w') = \inf\{F(w): w \in \overline{B}_V(v, r) \cap W\} = \inf \{F(w): w \in W\}.$$
