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!
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 $$F(w') = \inf\{F(w): w \in \overline{B}_V(v, r) \cap W\} = \inf \{F(w): w \in W\}.$$