Best approximation in a reflexive normed space 
Let $E$ be a reflexive normed space and let $\emptyset\neq K\subseteq E$ closed and convex. Show that then exists for every $x\in E$ a "best approximation" in $K$. Therefore a $y\in K$ with $\|x-y\|=d(x,K):=\inf\{\|x-z\|\colon z\in K\}$
Is additionaly $(E,\|\cdot\|)$ uniformly convex, then is the "best approximation" unique.

For the first part it is given as a hint, to show the existence of a bounded sequence $(y_n)_n\in K$ with $\|x-y_n\|\to d(x,K)$
I am stuck here. Can you help me out, or give me a hint?
Thanks in advance.
 A: Let $d := d(x, K)$. By definition of $\inf$, you can construct a minimizing sequence $(y_n)\subset K$ such that $\|x - y_n\| \to d$.
Namely, for every $n\in\mathbb{N}$ you can find a point $y_n\in K$ such that $d\leq \|x- y_n\| \leq d + 1/n$.
By construction, $(y_n) \in \overline{B}_{d+1}(x)$, hence $(y_n)$ is bounded.
Since $E$ is a reflexive Banach space, there exist a point $y_0\in E$ and a subsequence $(y_{n_j})$ of $(y_n)$ such that $y_{n_j} \rightharpoonup y_0$.
Moreover, since $K$ is convex and closed, you have that $y_0\in K$.
Let us prove that $\|x - y_0\| = d$.
Namely, by the lower semicontinuity of the norm with respect to the weak convergence, we have that
$$
d \leq \|x - y_0\| \leq \liminf_{j\to +\infty} \|x - y_{n_j}\| 
\leq \liminf_{j\to +\infty} \left(d + \frac{1}{n_j}\right) = d.
$$
Let us prove that the minimizer is unique if $E$ is uniformly convex.
Clearly, if $d=0$ (i.e., if $x\in K$), then the unique minimizer is $x$ itself.
Assume now that $x\not\in K$ (i.e. $d>0$), and assume by contradiction that there exist two points $y_0, y_1\in K$ such that
$$
\|x-y_0\| = \|x - y_1\| = d.
$$
Let us consider the normalized vectors
$$
\xi_i := \frac{x- y_i}{d},
\qquad i = 1,2.
$$
Setting $\epsilon := \|\xi_0 - \xi_1\| = \|y_0 - y_1\| / d > 0$, by the definition of uniform convexity there exists $\delta > 0$ such that
$$
\left\|\frac{\xi_0+\xi_1}{2}\right\| \leq 1 - \delta.
$$
On the other hand
$$
\left\|\frac{\xi_0+\xi_1}{2}\right\| =
\frac{1}{d} \left\|x - \frac{y_0 + y_1}{2}\right\|\geq 1,
$$
since $K$ is convex and so the point $(y_0+y_1)/2$ belongs to $K$.
