For a normed linear space $(X,||\cdot||)$ I want to show
a)For a continuous map $f : X \to \mathbb{R}$ the set $$f^{-1}(r):=\{x\in X: f(x)=r\} \subset X$$ is closed.
b) $$S(0,r)=\{x\in X: ||x||=r\}\subset X$$ is closed and bounded.
c) Provide an example that $S(0,r)$ is in general not compact.
I believe these questions lead on from eachother but I have a couple of things which confuse me.
My attempt at a):
Suppose we let $y$ be a limit point of $f^{-1}(r)$. So there is a sequence $\{y_n\}$ such that $y_n \in \{ x \in X : f(x)=r\}$ for all $n$ and $\lim_{n \to \infty} y_n = y$. Since $f$ is continuous we then have $f(y)= \lim_{n \to \infty}f(y_n)=\lim_{n \to \infty} r= r.$ Hence $ y \in f^{-1}(r)$, so $f^{-1}$ contains all of its limit points and is closed.
I understand this is a proof and generally 'show that' questions have a more direct approach so perhaps this is the wrong approach. Any feedback on my attempt is much appreciated.
Also in regards to b) and c), as far as I understood a compact space is one which is closed and bounded. But then b) and c) would contradict eachother so is this not the case all the time?
a) Your approach is correct, but it's simpler to say that, since $\{r\}$ is closed and $f$ is continuous, $f^{-1}(r)$ is closed.
b) It is closed by a) and it is bounded because each of its elements has norm $r$.
c) If $X$ is, say, the space of all bounded sequences of real numbers and if the norm of a sequence $(a_n)_{n\in\mathbb N}$ is $\sup_{n\in\mathbb N}|a_n|$, then the set $S(0,r)$ is closed and bounded, but not compact.