My intuition for (1) is to approximate the smooth, compactly-supported functions with compactly-supported step functions.
This is backwards. You should approximate things with smooth compactly supported functions, not the other way around.
I think there is no way around using mollifiers; how else will you construct smooth functions? Here is my approach:
Let $\Omega$ be any open subset of $\mathbb R^n$. For $x\in\Omega$, let $d(x)$ be the distance from $x$ to $\partial\Omega$. Given $f\in L^2(\Omega)$, define
$$f_n(x) = \begin{cases} f(x)\quad &\text{ if }\ d(x)>1/n \\ 0\quad &\text{ if }\ d(x)\le 1/n \end{cases}$$
Note that $(f_n-f)^2$ converges to $0$ pointwise, and is dominated by $f^2$. By the dominated convergence theorem, $\|f_n-f\|_{L^2}\to 0$.
Next, approximate $f_n$ by $f_n*\phi_\epsilon$, where the mollifier $\phi_\epsilon$ is supported in a ball of radius $\epsilon<1/n$. This will ensure that the convolution is both smooth and compactly supported in $\Omega$. The $L^2$ convergence of mollified functions is a standard fact.
You're correct in principle. Note that your question is vacuous without stating an ambient space. I'm assuming that for uniform convergence, our ambient space is the space of bounded functions and for uniform convergence on compacta (or local uniform convergence as I prefer to call it), that the ambient space is the space of locally bounded functions.
Denote by $\|\cdot \|_{\infty}$ the uniform norm and by $\|\cdot\|_{\infty, N}$ the semi-norm induced by the uniform norm on $C(B[0,N])$ . Then, the $f_m\to f$ uniformly if and only if $\|f_m-f\|_{\infty}\to 0,$ whereas $f_m\to f$ locally uniformly if and only if $\|f_m-f\|_{\infty,N}\to 0$ for every $N$.
Furthermore, I'll suppress $\mathbb{R}^n$ from my notation and e.g. write $C$, $C_c$ and $C_0$ for the function spaces in question.
Let's first show that $C_c$ is dense in $C_0$ w.r.t. to $\|\cdot \|_{\infty}$. So let $f\in C_0$ and let $\xi_N$ be a continuous function such that $0\leq \xi_N(x)\leq 1$ for every $x$, $\xi_N\equiv 1$ on $B[0,1]$ and $\xi_{N}\equiv 0$ on $\mathbb{R}^n \setminus B[0,N+1]$. Then, $\xi_Nf\in C_c$ and
$$
\|\xi_Nf-f\|_{\infty}\leq \| 1_{\{|x|\geq N\}} f\|_{\infty},
$$
which goes to $0$, since $f\in C_0(\mathbb{R})$.
Now, we need to show that $C_0$ is a closed subset of the bounded functions in the topology of uniform convergence. Indeed, if $f_m\in C_0$ for every $m$ and $f_m\to f$ uniformly, then $f$ is, necessarily, continuous. Furthermore, given $\varepsilon>0$, pick $m$ so large that $\|f_m -f\|_{\infty}<\varepsilon/2$ and $N$ so large that $|f_m(x)|<\varepsilon/2$ whenever $\|x\|\geq N$. Then, for $\|x\|\geq N,$ we get
$$
|f(x)|\leq |f(x)-f_m(x)|+|f_m(x)|<\frac{\varepsilon}{2}+\|f-f_m\|_{\infty}<\varepsilon,
$$
implying that $f$ tends to $0$ as $\|x\|\to \infty$. Hence, $C_0$ is closed, and we get that the closure of $C_c$ is $C_0$.
Let's now show that $C_c$ is locally uniformly dense in $C$. Let $f$ be continuous and note that $\|\xi_N f-f\|_{\infty,M}=0$ for $M\geq N$, establishing that $\xi_Nf\to f$ locally uniformly. I'm assuming that you know that $C$ is closed in the space of locally bounded functions (e.g. that locally uniform limits of sequences of continuous functions are again continuous). Hence, the locally uniform closure of $C_c$ is $C$.
Best Answer
No. In fact, its completion is the space of continuous functions $\mathbb{R}^n\to\mathbb{R}^{m}$ that vanish at $\infty$ (that is, $f(x)\to 0$ as $\|x\|\to\infty$). Given any continuous function $f$ which vanishes at $\infty$, let $f_k$ be obtained by "cutting off" $f$ to be supported on a ball of radius $k$ around the origin (so, say, $f_k$ agrees with $f$ on a ball of radius $k-1$, and interpolates radially to $0$ between the sphere of radius $k-1$ and the sphere of radius $k$). Then $f_k\in E$ for each $k$ and $f_k$ converges uniformly to $f$, and it follows easily that $(f_k)$ is Cauchy in $E$ but has no limit in $E$ unless $f$ has compact support.
There is no metrizable topology which induces this notion of convergence. To prove this, for each $R$, choose a sequence $(\varphi_k^R)_k$ such that each $\varphi_k^R$ has a ball of radius $R$ as its support but $\varphi_k^R\to 0$ uniformly. Then each of these sequences $(\varphi_k^R)$ converges to $0$ in Maggi's sense. If this notion of convergence came from a metric, we could choose for each $R$ a $k_R$ such that $\varphi_{k_R}^R$ is within $1/R$ of $0$ with respect to the metric, and then $\varphi_{k_R}^R$ would converge to $0$ as $R\to\infty$. But this is impossible since convergent sequences are required to have some uniform compact support.
However, this convergence does come from a natural topology. For each compact $K\subset\mathbb{R}^n$, let $E_K$ be the subspace of $E$ consisting of functions with support contained in $K$. Say that a set $U\subseteq E$ is a basic neighborhood of $0$ if $U$ is balanced and convex, and $U\cap E_K$ is an open neighborhood of $0$ in $E_K$ with respect to the sup norm on $E_K$ for each $K$. Finally, say that a set $U\subseteq E$ is open if it is a union of translates of basic neighborhoods of $0$.
It can be shown that this topology makes $E$ as the colimit of the spaces $E_K$ with their sup norms in the category of locally convex topological vector spaces. That is, it is the finest topology on $E$ which makes the inclusion maps $E_K\to E$ all continuous and makes $E$ a locally convex topological vector space. It is then clear that any sequence which converges in Maggi's sense converges in this topology, since such a sequence will converge in $E_K$ with respect to the sup norm.
Proving the converse (every sequence convergent in this topology converges in Maggi's sense) is a little messier, but here's the idea. Suppose $(f_k)$ is a sequence of functions in $E$ such that the supports of the $f_k$ are not contained in any fixed compact set. Then, passing to a subsequence of $(f_k)$ if necessary, we can find a sequence $(x_k)$ going to $\infty$ in $\mathbb{R}^n$ such that $f_k(x_k)\neq 0$ for each $k$. Now let $U$ be the set of $f\in E$ such that $|f(x_k)|<|f_k(x_k)|$ for each $k$. Then $U$ is a basic neighborhood of $0$ in $E$ (this uses the fact that any $K$ contains only finitely many of the $x_k$). Since $f_k\not\in U$ for all $k$, this means $(f_k)$ cannot converge to $0$.