Let $X$ be a norm space, $S$ be a subset of $X$, for each $x\in X$, denote the set of projection from $x$ to $S$ by $\Pi(x; S):=\{s\in S: \ \|x-s\| =d(x;S)\}$, where $d(x;S):=\inf\{\|x-s\|: \ s\in S\}$.
It is well-known that $\Pi(x;S)$ is nonempty if one of the following assertions holds: (i) $X$ is finite dimensional and $S$ is closed. (ii) $X$ is reflexive and $S$ is weakly closed. Thus, if $X$ is reflexive and $S$ is closed and convex then $\Pi(x; S) \neq \emptyset$, too.
Now, I want to show that the assumption about reflexivity of $X$ is important in the case (ii), my teacher give a counter example below: Consider $\ell^1$ and take $S$ to be the closed convex hull of the set $\{\frac{n+1}{n}e_n: n\in \mathbb{N}\}$ where $e_n=(0, \ldots, 1_n, 0, \ldots)$. I don't know how to verify that $\Pi(0; S)=\emptyset.$ Can anyone help me? Thanks in advance.
Let's start from a finite dimensional case. In $\mathbb R^2$ the convex hull of $e_1,e_2$ is a segment, each point of which has $l^1$-distance from $0$ exactly $1$. In $\mathbb R^3$, the convex hull of the basis $e_1,e_2,e_3$ is a triangle, each point of which, has $l^1$-distance from $0$ exactly $1$. And so on... the same argument holds for the space $l^1$, when you take the convex hull of $e_n$. In this case the $l^1$-distance from $S$ to $0$ is constant $1$ and $\Pi(0;S)=S$.
Now, your teacher, pushed the points $e_n$ a little away from zero, by an amount $(n+1)/n$ which tends to $1$. Therefore the infimum of the $l^1$-distances from points of $S$ to $0$ is $1$, but there is no point that realizes such distance.