Closed point problems in $l$-spaces, namely $ l^1$ and $l^\inf$ a) Give an example of a closed convex set $C$ and point $x$ in $l^\infty$ such that the closed point in $C$ to $x$ is not unique.
Solution: So I was thinking that a closed convex set would be the union of two closed balls in $l^\infty$ with a shared face as these will be cubes which are convex structures. 
Let $C = \bar B(x,1) \cup \bar B(y,1)$; where $x = (0,0,...)$ and $y=(2,2,..)$
Consider $x= (-2,4,-2,4,-2,...)$ then $x \in l^\infty$
Then $\left\lVert x-(1,1,1,...) \right\rVert_\infty = \left\lVert x-(1,1/2,1,1/2...) \right\rVert_\infty = 3$ is are both the closest points to x where $\left\lVert \cdot \right\rVert_\infty = sup_{n \in \mathbb N} \lvert c_n \rvert $
I feel like this isn't a sufficient answer so I'm wondering what else I might need to show.
b) Show there is no closest point to $0$ in $C$ = {${x \in l: f(x) = 1}$} where $f:l^1 \rightarrow \mathbb R$ and $f(x) = \sum_{n=1}^\infty (1-1/n)x_n$
I'm not so sure how to go about this.
Suppose there was $y \in C$ such that $\lvert f(x) \rvert = \left\lVert x \right\rVert_1$ then this would be the closest point to $0$ in $C$ and so this would mean $\left\lVert x \right\rVert_1 = \sum_{n=1}^\infty \lvert x_n \rvert = 1$.
But in that case we have $\sum_{n=1}^\infty \lvert x_n \rvert = \lvert \sum_{n=1}^\infty (1-1/n)x_n \rvert$
And then I'd argue that this equality is never satisfied?
 A: I have the impression that you are looking in the right direction, but your ideas are not really getting anywhere.
Regarding a), your set $C = \overline B(x,1) \cup \overline B(y,1)$ with $x = (0, 0, \ldots)$ and $y = (2, 2, \ldots)$ is not convex, because the two cubes don't share a face. Consider for example the points $r = (1,0,0,\ldots)$ and $s = (2,1,2,2,\ldots).$ We have $r\in \overline B(x,1)$ and $s \in \overline B(y,1).$ Put
$$
t = \frac12 (a+b) = (\frac32,\frac12,1,1,\ldots).
$$
We have $t\notin \overline B(x,1)$ because of the first coordinate, and $t\notin \overline B(y,1)$ because of the second coordinate, and thus $C$ is not convex.
We could construct an example using two cubes that share a face, but it's simpler to use just one cube. Put
$$
E = \overline B(x,1).
$$
Then $E$ is closed and convex. Put
$$
z = (3, 0, 0, \ldots).
$$
Please convince yourself that $dist(E,z) = 2$ and
$$
\{u = (u_1,u_2,\ldots) \in E | \|u-z\|_\infty = 2\} = \{(1,u_2,u_3,\ldots) | |u_j| \leq 1\ for\ j\in \{2,3,\ldots\}\}.
$$
This is a face of the cube $E,$ and certainly contains more than one element.
On to b). Let's put
$$
\varphi_n =1-\frac1n\quad for\ n\in \{1,2,\ldots\}.
$$
Then we consider the space $l^1$ and the linear map
$$
f:l^1\rightarrow \mathbb R, x = (x_1,x_2,\ldots) \mapsto \sum_{n=1}^\infty \varphi_n x_n.
$$
$f$ is nonzero, and it's "easy to see" that $f$ is continuous (use the fact that $0\leq\varphi_n \leq 1).$ This implies that the set
$$
G = f^{-1}(\{1\}) = \{ x \in l^1 | f(x) = 1\}
$$
is a closed, affine hyperplane ("affine" means "not passing through $0$"). Since affine hyperplanes are convex, $G$ is convex. Since $G$ is closed and $0\notin G,$ we must have
$$
dist(G,0) = \inf\{x\in G|\|x\|_1\} > 0.
$$
I have no idea about the exact value of that distance.
Now you want to argue that if we had a $y\in G$ with
$$
dist(0,y) = \|y\|_1 = f(y) = 1,
$$
we would get a contradiction. However, I can't see how to prove the existence of such a $y.$ Here's another argument.
Assume that we have a $u = (u_1,u_2,\ldots) \in G$ nearest to $0,$ i.e.
$$
dist(u,0) = \|u\|_1 = dist(G,0).
$$
Put
$$
v = (0,\frac{\varphi_1}{\varphi_2}u_1,\frac{\varphi_2}{\varphi_3}u_2,\ldots).
$$
From
$$
0\leq \varphi_n < \varphi_{n+1}\quad for\ n\in\{1,2,\ldots\},
$$
and $v\neq 0,$ we can infer
$$
\|v\|_1 < \|u\|_1.
$$
Note the strict inequality. Moreover, we compute
$$
f(v) = \sum_{n = 1}^\infty \varphi_{n+1}\frac{\varphi_n}{\varphi_{n+1}}u_n = f(u) = 1,
$$
and so
$$
v \in G.
$$
Now we have
$$
dist(G,0) \leq dist(v,0)= \|v\|_1 < \|u\|_1 = dist(u,0) = dist(G,0).
$$
That's a contradiction. So, in fact, there's no point in $G$ nearest to $0,$ i.e. of minimal norm, althoug (as noted above), $dist(G,0) > 0.$
Note: if you assume the existence of $u,$ there are other ways to arrive at a contradiction besides the "weighted shift" used above. For example, it's enough to "appropriately manipulate" just two coordinates of $u.$
