Ok this is partially so I know what you have in mind without revealing too much

@Daminark Let $v_n$ be a sequence of vectors s.t. $d(v_n, w)$ approaches its infimum. Of course $d(v_n, w) + d(w, v_m) \geq d(v_n, v_m)$, and the LHS can be bounded by epsilon by choosing $n, m > N$. So $v_n$ is Cauchy, and since $V$ is closed (and I assume we're working somewhere complete, right?) has a limit $v$ achieving the minimum distance. Then I claim $v - w$ is perpendicular to $V$.

$\|v-w+tv'\|^2 = \|v-w\|^2 + t 2\langle v', v-w\rangle + t^2\|v'\|^2$. We know this is strictly greater than $\|v-w\|^2$ by assumption. So you have $2t\langle v', v-w\rangle > -t^2\|v'\|^2$. Divide by $t$ and take a limit as $t \to 0$ to see that $\langle v', v-w\rangle \geq 0$. The same argument for $-v'$ gets what we want.

Once $w-v$ is perpendicular to $V$ we have our contradiction: $\langle w - v, w - v\rangle = \langle w, w-v\rangle - \langle v, w-v\rangle$. $w - v \in V^{\perp \perp}$, so both of those terms vanish, and $w = v$.

I'm not expecting the first argument - throwing the triangle inequality and completeness out there are a little technical - but I want him to start on the approach.

The only thing with the second argument is the $t$, which is tricky.

Hmm, the way I saw it was basically this

So let $E$ be a closed, convex subset of a Hilbert space $\mathcal{H}$, then there's a unique $x_0 \in E$ such that $\|x_0\| = dist(0,E)$

Sure, argument should be the same idea, more or less. You try and get as close as you can to 0.

Actually I expect the argument will be identical.

The only change should be where you invoke convexity: if $v_n, v_m$ are two points far enough down the sequence, halfway between them is $(v_n - v_m)/2$, whose distance to $0$ you can bound. Thus Cauchy.

So we have that $\mu = dist(0,E) = \inf_{v\in E} \|v\|$. Take a minimizing sequence, $\{x_n\}$ and apply the parallelogram law immediately. Then $\|\frac{x_n - x_m}{2}\|^2 = \frac{1}{2})(\|x_n\|^2 + \|x_m\|^2) - \|\frac{x_n + x_m}{2}\|^2$. By convexity, the last term is in $E$, and for big enough $m,n$, we can bound this by $2\mu\epsilon + \epsilon^2$

Yeah, same argument so far.

This is why I'm trying to guide him slightly toward parallelogram law

If he can come to this, then he'll be in good shape to whip out some chalk and write

I don't really think the parallelogram law is easier than the triangle inequality...

Perhaps, I dunno