$C(S)$ is never compact for any $S$! Indeed it is a normed vector space and no non-zero normed vector space is compact:
$$\bigcup_{n\in\mathbb N} B_n(0)=V$$
but $V$ definitely isn't $\bigcup_{m=1}^M B_{n_m}(0)=B_{n_M}(0)$, where $n_1<n_2<...<n_M$ is a finite subsequence of $\mathbb N$. (Here $B_\epsilon(x)$ is the epsilon ball around $x$.)
However, for a compact $X$ you have that $C(X)$ is separable, this is an application of Stone-Weierstrass, you can read about it here.
Our situation is the following:
$$BL(S)\cong BL(\tilde S) \hookrightarrow C(\tilde S)$$
where the inclusion into $C(\tilde S)$ is isometric. So basically you can identify $BL(S)$ with a subset of $C(\tilde S)$, which is separable. But any subset of a separable metric space is itself separable. The proof goes like this:
Let $\{x_n\mid n\in\mathbb N\}$ be a dense subset of $C(\tilde S)$. For each $n,m\in\mathbb N$ choose an element $y_{n,m}\in BL(S)$ so that
$$y_{n,m}\in\{y\in BL(S)\mid d(x_n,y)<\frac1m\}$$
provided that set is non-empty of course (if it is empty set $y_{n,m}=0$). The set
$$\{y_{n,m}\mid (n,m)\in\mathbb N^2\}$$
is a countable subset of $BL(S)$.
It is also dense, for if $y\in BL(S)\subset C(\tilde S)$ and $\epsilon>0$ you have an $x_n$ and an $m$ so that $\epsilon>\frac2m$ and $B_{1/{m}}(x_n)$ contains $y$ (since $\{x_n\}$ is dense). But then $y_{n,m}$ has distance $<\frac1m$ from $x_n$ and thus distance $<\frac2m<\epsilon$ from $y$. Thus every open set around $y$ contains an element from $y_{n,m}$.
