For a topological space $X$, let $C_0(X)$ denote the set of continuous functions $$ f:X\to \mathbb{C} $$ such that for all $\epsilon>0$, the set $\{x\in X\mid |f(x)|\geq \epsilon\}$ is compact.
(a) Show that $C_0(X)$ is complete with respect to the norm $\|f\|_\infty:=\sup_{x\in X}|f(x)|$
For (a) I did the standard thing and took $(f_n)$ to be a Cauchy sequence in $C_0(X)$.
For each $x\in X$, $(f_n(x))_{n\in\mathbb{N}}$ is Cauchy and therefore converges to a point, say f(x) (as $\mathbb{C}$ is complete). My claim is that the limit of $(f_n)$ is the pointwise limit of the $f_n(x)$'s.
Fix $\epsilon>0$.
Since $(f_n)$ is a Cauchy, by definition there exists $N\in\mathbb{N}$ such that for all $m,n>N$, $|f_m(x)-f_n(x)|<\epsilon.$ So taking the limit as $n\to\infty$ we get that $|f_m(x)-f(x)|\leq\epsilon.$ Since this is independent of $x$ we actually have $\sup\limits_{x\in X}|f_m(x)-f(x)|=\|f_m-f\|_\infty\leq \epsilon$. Hence $f_m\to f$.
This is where I'm a little unsure. Can I use the fact that uniform limit of continuous function is continuous and how do I know for all $\epsilon>0$, the set $\{x\in X\mid |f(x)|\geq \epsilon\}$ is compact?
For part $(b)$ and $(c)$ I'm having trouble coming up with the isometries.
Any help is appreciated.