I don't think you need to assume that $X$ is locally compact:
Let $f\in C_{0}\left(X\right)$ and $\varepsilon>0$. Note that $U:=\left\{ x\in X\,\mid\,\left|f\left(x\right)\right|>\frac{\varepsilon}{2}\right\} $
is open with closure $\overline{U}\subset\left\{ x\in X\,\mid\,\left|f\left(x\right)\right|\geq\frac{\varepsilon}{2}\right\} $,
which is compact as a closed subset of a compact set. Furthermore,
$K:=\left\{ x\in X\,|\,\left|f\left(x\right)\right|\geq\varepsilon\right\} $
is a compact subset of $U$, so that the Urysohn-Lemma (applied to
the compact Hausdorff space $\overline{U}$, with $U\subset\overline{U}$
open and $K\subset U$ compact) yields $\varphi\in C(\overline{U})$
with $\varphi|_{K}\equiv1$, $0\leq\varphi\leq1$ and $\operatorname{supp}\varphi\subset U$,
where the closure for the support is taken in $\overline{U}$.
Now,
define
$$
g:X\to\left[0,1\right],x\mapsto\begin{cases}
0, & x\in X\setminus U,\\
\varphi\left(x\right), & x\in\overline{U}.
\end{cases}
$$
Note that $X\setminus U$ and $\overline{U}$ are both closed, so
that $g$ is continuous by the gluing lemma (note that $g$ is well-defined,
since for $x\in\left(X\setminus U\right)\cap\overline{U}$, we have
$\varphi\left(x\right)=0$ since $\operatorname{supp}\varphi\subset U$.
Thus, $g$ has compact support (in fact $\operatorname{supp}g=\operatorname{supp}\varphi$),
so that the same is true of $\tilde{f}:=g\cdot f\in C_{c}\left(X\right)$.
It is not hard to see $\left\Vert \tilde{f}-f\right\Vert _{\sup}\leq\varepsilon$.
$C_0(X)$ generates the same $\sigma$-algebra as $C_c(X)$ (and thus the same $\sigma$-algebra as the compact $G_\delta$ sets). This follows from the fact that every element of $C_0(X)$ is the pointwise (in fact, uniform) limit of a sequence in $C_c(X)$: given $f\in C_0(X)$ and $\epsilon>0$, there is a compact set $K$ such that $|f|<\epsilon$ outside of $K$, and then we can extend $f|_K$ to a function with compact support that stays no larger than $\epsilon$ outside of $K$. Since a limit of a sequence of measurable functions is measurable, this means every function in $C_0(X)$ is $\sigma(C_c(X))$-measurable.
Best Answer
Due to Hahn-Banach, the closure of a subspace in the weak topology of a normed space is equal to the norm closure. Then it is easy to check (without using the Stone-Cech compactification as in Eric Wofsey's answer) that $\overline{C_c(X)}=C_0(X)$ (the space of all continuous functions which, for all $\varepsilon>0$, are less than $\varepsilon$ outside some compact set). What you technically need is normality of locally compact Hausdorff spaces to find cut-off fuctions.