Can someone help me with the following question, I know that for a subset to be compact it must be closed and bounded, but i'm not sure how to show that for this case.
Let $(X,\|\cdot\|)$ be a normed linear space
$$S(0,r)=\{x\in X:\|x\|=r\}\subset X $$
Show by an example what $S(0,r)$ is in general not compact.
On
Not all closed bounded subsets of normed space are compact. In particular
If a normed space is not metrically complete (e.g. it has some Cauchy sequences that do not converge within the space) then it has some closed and bounded subsets that are not compact.
If a normed space is infinite-dimensional it can have closed bounded subsets that are not compact.
On of the simplest examples is the Hilbert space $\ell^2$ consisting of all infinite sequences $(c_k)_{k=1}^\infty$ for which $\sum_{k=1}^\infty |c_k|^2 < \infty.$ In that space the set $\{ (0,\ldots,0,1,0,0,\ldots), \ldots \}$ of all vectors having a single component equal to $1$ and the rest equal to $0$ is a closed bounded set that is not compact. To see that, just cover each such point with an open ball of radius $1/10,$ and there you have an open cover with no finite subcover. And it is a subset of $S(0,1),$ which is therefore a closed bounded set that is not compact.
The compactness criterion (being closed and bounded) holds for subsets of $\mathbb{R}^n$, but not in general.
As an example, consider the space $X$ of all bounded sequences of ral numbers, with the norm $\bigl\|(x_n)_{n\in\mathbb N}\bigr\|=\sup_{n\in\mathbb N}|x_n|$. Tnen $S(0,r)$ is not compact (although closed and bounded). In fact, consider the sequence $\bigl(x(n)\bigr)_{n\in\mathbb N}$ of elements of $X$ thus defined:$$x(n)_k=\begin{cases}r&\text{ if }k=n\\0&\text{ otherwise.}\end{cases}$$This sequence has no convergent subsequence, since$$(\forall m,n\in\mathbb{N}):m\neq n\implies\bigl\|x(m)-x(n)\bigr\|=r.$$ Therefore, $S(0,r)$ is not compact.