Show that S is non-compact and deduce further that the closed unit ball in X is non-compact. I have an exercise I am not sure how to solve
The exercise is:
Let X be an infinite dimensional normed vector space over $\mathbb{K}$ where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$. Define the unit sphere $S=\{x \in X : \|x\|=1\}$
Show that S is non-compact and deduce further that the closed unit ball in X is non-compact.
I am not sure how to show that S is non-compact, but for the second part where I have to deduce that the closed unit ball in X is non-compact (I denote closed unit ball in X by B)  I am thinking that I will show B is compact $\Rightarrow$ X is finite.
I assume that B is compact and then since S is a closed subset of B, then S will be compact and then I can use the first part to say that X may be finite. So now I have shown that B is compact $\Rightarrow$ X is finite, hence if X is infinite then B is non-compact. But I have troubles showing that S is non-compact, so I was hoping that someone could help me with that. We have already shown following:

 A: Your Problem 3, (d) implies that $S$ is not compact.
Indeed, assume that $S$ is compact and for any $x\in S$ consider $B_x=\{v\in X\ |\ \lVert x-v\rVert < 1/2\}$. Then $\{B_x\}_{x\in S}$ is an open covering of $S$ and by compactness it has to contain a finite subcover $\{B_{x_1}, \ldots, B_{x_n}\}$. But then $\{\overline{B_{x_1}},\ldots, \overline{B_{x_n}}\}$ is a closed ball covering of $S$ and none of them contains $0$, contradicting Problem 3, (d).
Finally the closed ball cannot be compact as well because $S$ is its closed subset. And a closed subset of a compact space is compact.
Also, please do you use full "finite dimensional" and "infinite dimensional" instead of "finite" and "infinite" to avoid confusion. It took me some time to figure out that you are not talking about literally finite sets.
A: $(1).$ Suppose $Y$ is a closed vector-subspace of $X$ and $x\in X$ \ $Y.$ Let $d(x,Y)=\inf_{y\in Y}\|x-y\|$. Take $z\in Y$ with $$(*)...\quad \|x-z\|<2d(x,Y).$$ Let $w=(x-z)/\|x-z\|.$ Then we have $$(**)...\quad d(w, Y)=\inf_{y\in Y}\|w-y\|=\|x-z\|^{-1}\inf_{y\in Y}\|\,(x-(z+y\|x-z\|)\,\|.$$ Now $\{z+y\|x-z\|:y\in Y\}=Y.$ So $\inf_{y\in Y} \|(x-(z+y\|x-z\|)\|=\inf_{y\in Y}\|x-y\|=d(x,Y).$ So from $(*)$ & $(**)$ we have $$d(w,Y)=\|x-z\|^{-1}d(x,Y)>\|x-z\|^{-1}\|x-z\|/2=1/2.$$
So if $Y$ is a closed vector-subspace of $X$ and $Y\ne X$ then there exists $w\in X$ \ $Y$ with $\|w\|=1$ and $d(w,Y)>1/2.$
$(2).$ Prove that if $Y$ is a finite-dimensional vector-subspace of $X$ then $Y$ is closed.
$(3).$ If $X$ is infinite-dimensional: Take $y_1\in X$  with $\|y_1\|=1.$ Recursively for $n\in \Bbb N$ let $Y_n=Span(y_1,...,y_n)$ and by $(1)$ & $(2)$ take $y_{n+1}\in X$ \ $Y_n$ with $\|y_{n+1}\|=1$ and $d(y_{n+1}, Y_n)>1/2.$
Let $T=\{y_n:n\in \Bbb N\}.$ Then $T$ is an infinite set and $\|y-y'\|>1/2$ for any  distinct $y,y'\in T.$ So $T$ is an infinite closed discrete topological-subspace of $X.$
Consider $X, S$, and the closed unit ball $B$ as metric spaces with the metric $e(u,v)=\|u-v\|.$ Then $T$ is an infinite closed discrete topological-subspace of $S$ and of $B.$ Any metric space with such a topological-subspace cannot be compact.
[... Any subset of a closed discrete topological-subspace is also closed. So if $U_n=X\setminus \{y_j:j\ge n\}$ then $\{U_n:n\in\Bbb N\}$ is an open cover of $S$ (and of $B$) with no finite subcover...].
Remarks: In $(1)$ we could take $z\in Y$ with $\|x-z\|$ arbitrarily close to $d(x,Y)$ and hence $d(w,Y)$ can be arbitrarily close to $1.$ And if $Y$ is finite-dimensional we can show there exists $z\in Y$ with $d(x,Y)=\|x-z\|$ and hence $d(w, Y)=1$.
