Let $C\subset E$ be a convex subspace and $a\in E\setminus C$. The point in $C$ closest to $a$ is unique. 
Let $E$ be a vector space with a norm $|\cdot|$ induced by an inner product, $C\subset E$ a convex set, and $a\in E\setminus C$. Show that if there are $x_0,x_1 \in C$ such that $|a-x_0|\leq |a-x|$ and $|a-x_1|\leq|a-x|$ for every $x\in C$, then $x_0 = x_1$.

Picturing a ball, the only point with such property is the one closest to $a$, which is unique, so the question makes sense. However, I haven't been able to use convexity of $C$ to finish the proof.
I found similar questions in the context of real analysis, but couldn't abstract the logic.
 A: The geometric proof others have suggested would also be the way I'd recommend proceeding. However there does exists a more analysis based proof. Let $d = \inf_{x \in C} \lvert a - x \rvert$ and $(x_i) \subset C$ be any sequence such that $\lvert a - x_i \rvert \to d$ as $i \to \infty$. We aim to prove $(x_i)$ is a Cauchy sequence. Since the norm arises from an inner product $\langle \cdot, \cdot \rangle$, we know the norm satisfies the parallelogram identity:
$$ \lvert v + w \rvert^2 + \lvert v - w \rvert^2 = 2\lvert v \rvert^2 + 2\lvert w \rvert^2 $$
Plugging in $v = a - x_i$ and $w = a - x_j$ and rearranging gives that
$$ \lvert x_i - x_j \rvert^2 = 2\lvert a - x_i \rvert^2 + 2 \lvert a - x_j \rvert^2 - 4 \lvert a - \tfrac{1}{2}(x_i + x_j) \rvert^2$$
Since $C$ is convex we know $\frac{1}{2}(x_i + x_j) \in C$ therefore $\lvert a - \frac{1}{2}(x_i + x_j) \rvert \geq d$. Moreover since $\lvert a - x_i \rvert \to d$ as $i \to \infty$ we can find $N$ such that for all $i \geq N$ we have $\lvert a - x_i \rvert^2 < d^2 + \epsilon$. Then for all $i, j \geq N$ we have
$$ \lvert x_i - x_j \rvert^2 < 2(d^2 + \epsilon) + 2(d^2 + \epsilon) - 4d^2 = 4\epsilon $$
Therefore $(x_i)$ is Cauchy.
Now suppose that $\lvert a - x \rvert = d$ is achieved for two points in $C$, say $v_0$ and $v_1$. Then we can take $(x_i)$ to be the alternating sequence $v_0, v_1, v_0, v_1, \ldots$ and the above shows this sequence is Cauchy. So in fact $v_0 = v_1$.
This proof has some value because if we further assume $C$ to be complete (in particular $C$ is complete if $E$ is a Hilbert space and $C$ is closed), we know any Cauchy sequence has a limit. Therefore taking any $(x_i) \subset C$ such that $\lvert a - x_i \rvert \to d$, we know $x_i \to x$ for some $x \in C$ as $i \to \infty$. Moroever by continuity of the norm $\lvert a - x \rvert = d$. Therefore if $C$ is complete and convex, we can always achieve the minimum distance from $C$ to any point.
A: Just a hint: If $x_0 \ne x_1$ are both at distance $d,$ then one of the points interior to the segment joining $x_0$ to $x_1$ should be at distance less than $d.$
