Definition: Let $f:\mathbb{R}^N\to\mathbb{R}$ be a function. Consider the family $\omega_i$ of all open sets on $\mathbb{R}^N$ such that for each $i$, $f=0$ a.e. in $\omega_i$, then $f=0$ a.e. on $\omega=\cup\omega_i$ and we define the support of $f$ by $$\operatorname{supp}(f)=\mathbb{R}^N\setminus\omega$$
Suppose that $f,g\in C^\infty(\mathbb{R})\cap L^1(\mathbb{R})$. We can prove that $\operatorname{supp}(f\ast g)\subset\overline{\operatorname{supp}(f)+\operatorname{supp}(g)}$.
Take any $x\in\mathbb{R}$ and note that $$(f\ast g)(x)=\int_\mathbb{R}f(x-y)g(y)=\int_{(x-\operatorname{supp}(f))\cap \operatorname{supp}(g)}f(x-y)g(y)$$
For $x\notin \operatorname{supp}(f)+\operatorname{supp}(g)$, we have that $(x-\operatorname{supp}(f))\cap \operatorname{supp}(g)=\emptyset$ and then $(f\ast g)(x)=0$. It follows that if $x\in (\operatorname{supp}(f) + \operatorname{supp}(g))^c$ then $$(f\ast g)(x)=0$$
Therefore $$\operatorname{supp}(f\ast g)\subset\overline{\operatorname{supp}(f)+\operatorname{supp}(g)}$$
Now let's prove that the equality is not necessary. Consider the functions in $\mathbb{R}$ defined by $f(x)=x\chi_{[-1,1]}(x)$ and $g(x)=\chi_{[-2,2]}(x)$, where $\chi_A$ is the characteristic function of the set $A$.
Note that in this case $\overline{\operatorname{supp}(f)+\operatorname{supp}( g)}=[-3,3]$, however, the interval $(-1,1)$ does not belong to the support of $f\star g$ (which is equal to $[-3,-1]\cup [1,3]$).
Remark 1: The proof I gave here can be found in Brezis's book of Functional Analysis.
Remark 2: I don't know a expliclty characterization of the support of the convolution, but by the given formula, you can see that if the two functions has compact support, then does the convolution.
Update: I have corrected some errors in the text.
Your idea is good, and you must only take a little more care for it to work. As you said, if $K$ is a compact metric space, then $C(K)$ is separable.
Now, suppose that $X$ is is a locally compact, $\sigma$-compact metric space. You can find a sequence $\left\{K_n\right\}_{n\in\mathbb{N}}$ of compact subsets of $X$ satisfying:
$K_n\subseteq\text{int}K_{n+1}$ for every $n$;
$X=\bigcup_{n\in\mathbb{N}}K_n$.
For every $n$, let $C_n=\left\{f\in C_c(X):\text{supp}f\subseteq K_n\right\}$. Notice that $C_c(X)=\bigcup_{n\in\mathbb{N}}C_n$, so it is sufficient to show that each $C_n$ is separable.
Fixed $n$, consider the function $R_{K_n}:C_n\rightarrow C(K_n)$, $f\mapsto f|_{K_n}$. This is a linear isometry (not necessarily surjective), so $R_{K_n}(C_n)$ is a subspace of the separable space $C(K_n)$, so it is also separable, hence $C_n$ is also separable.
Best Answer
Not in general. Let Consider the following function, where $n = 1,2,...$:
$$f(x) = \begin{cases} 0 & \text{ if } x < 0 \\ \frac{1}{n^3}& \text{ if }\frac{n^2-n}{2} \leq x < \frac{n^2+n}{2}\end{cases}$$
This function is $L^1$ with total integral $\frac{\pi^2}{6}$. However, it is constant and non-zero on arbitrarily long intervals. Thus for any $\phi$ as specified, there is sufficiently large $N$, such that for all $n >N$, there is some $x$ in the interval $\frac{n^2-n}{2} < x<\frac{n^2+n}{n}$ with $(f * \phi)(x) = \frac{1}{n^3}>0$.