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.

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.
