Can you tell me if my proof is correct? Thank you!
Claim: $C_0(X)$ is a closed subspace of $C_b(X)$
Proof: We have to show that $C_0(X)$ contains all of its limit points. Let $f(x)$ be a limit point of it, then we have a sequence $f_n$ converging to it (in $\|\cdot\|_\infty$) hence $f_n$ is a Cauchy sequence (with respect to $\|\cdot\|_\infty$). Of course, $f$ is continuous since it's the uniform limit of a sequence of continuous functions. So we only have to show that $f(x)$ vanishes at $\pm \infty$ (apparently, $X \subset \mathbb R$? See here for a definition of $C_0(X)$. I'd have thought it's more general but then what does $x \to \infty$ mean in a general topological space?)
Let $\varepsilon>0$. Let $N$ be such that $m,n \geq N$ implies $\|f_n - f_N\|_\infty \leq \varepsilon/2$. Then
$ |f(x)| \leq |f(x) - f_N(x)| + |f_N(x)| \leq \varepsilon$
for $x \in X \setminus K_N$ where $|f_N(x)| \leq \varepsilon/2$ outside some $K_N$, $K_N$ compact. Also, for all $x$, $|f(x) - f_N(x)| = \lim_{m \to \infty} |f_m(x) - f_N(x)| \leq \varepsilon/2$ since $|f_m(x) - f_N(x)| \leq \|f_m - f_N\|_\infty \leq \varepsilon / 2$ for all $m > N$.