In a Reflexive banach space, given a closed convex set $C$ and some point $y$, there is a point in $C$, of minimal distance to $y$

A subset $C$ of X which has the property that for every $x in X$, the distance $operatorname{dist}(x,C)$ is attained (i.e there exists a $c in C$ s.t $||x-c||=operatorname{dist}(x,C)$) is called proximinal.

We shall show that if $C$ is a closed and bounded subset of a reflexive space $X$, then $C$ is proximinal. Let $x in X$ and $d=operatorname{dist(x,C}):= inf { ||{x-c}|| colon c in C}$. By the definition of the infimum, for all $n in mathbb N$, we can find a $ z_n in B_{d+1/n}(x) cap C := { c in C colon ||x-c|| leq d+1/n}$ (so the intersection is not empty). Define the sequence of subsets $(A_n)_n subset 2^X$, by $A_n = B_{d+1/n}(x) cap C$. Notice that $B_{d+1} supset A_1 supset A_2 supset dots $ and $B_{d+1}$ is weakly compact (since X is reflexive). Furthermore, one can easily check that the famlily $(A_n)_n$ has the finite intersection property and so, we can find a $z in bigcap_{n in mathbb N} B_{d+1/n}cap C$. The point $z$ is the required.

In fact, it can be shown that a normed space X is reflexive if and only if every closed and convex subset of X is proximinal. (The opposite direction is due to James' theorem.)

You are on the right track. Since $C$ is not only closed but also convex, it is weakly closed. The weak limit of the sequence $(z_n)_{n in mathbb{N}}$, lets call it $z$, therefore satisfies $z in C$. Can you show that $z$ indeed minimizes the distance using that the norm is (sequentially) weakly lower semi-continuous?