Let $j(x)$ be any positive infinitely differentiable function with support in $(-1, 1)$ so that $$\int_{-\infty}^\infty j(x) dx = 1.$$ Define $$j_\epsilon(x) = \epsilon^{-1} j(x/\epsilon).$$ If $f \in C^1_c(\mathbb{R})$, the set of continuously differentiable functions with compact support, set $$f_\epsilon(x) = \int_{-\infty}^\infty j_\epsilon(x-t) f(t) dt.$$
In Reed & Simon's textbook on functional analysis they first show that $$|f_\epsilon(x) - f(x)| \leq \sup_{\{t : |x-t| \leq \epsilon\}} |f(t) - f(x)|$$ and then state:
Since $f$ has compact support, it is uniformly continuous which implies that $f_\epsilon \rightarrow f$ uniformly. Since the $f_\epsilon$ have compact support in a fixed compact set, $f_\epsilon \rightarrow f$ in $L^2(\mathbb{R})$.
How does $f$ being uniformly continuous imply that $f_\epsilon \rightarrow f$ uniformly? Furthermore, how does $f_\epsilon$ having support in a fixed compact set imply the $L^2(\mathbb{R})$ convergence of $f_\epsilon \rightarrow f$?
Let's tackle the first assertion. By using the fact that $$\int_\mathbb{R} j_\epsilon = \int_\mathbb{R} j = 1,$$ we have that $$|f_\epsilon(x)-f(x)| \leq \int_\mathbb{R} j_\epsilon(t)|f(x-t)-f(x)| \, dt = \int_\mathbb{R} j(t)|f(x-\epsilon t)-f(x)| \, dt. \qquad (1)$$ Since $f$ is uniformly continuous, for any fixed but arbitrary $\eta > 0$ there is some $\delta > 0$ such that $$|f(x -\epsilon t) -f(x)| < \eta \quad \text{ for all } |t| < \frac{\delta}{\epsilon}. \qquad (2)$$ Now choose $\epsilon_1 > 0$ such that $\operatorname{supp} j \subseteq [-1,1] \subseteq [-\delta / \epsilon_1 , \delta / \epsilon_1]$. Then, for any $\epsilon < \epsilon_1$, we go back to $(1)$ and use $(2)$ to conclude that \begin{align} |f_\epsilon(x)-f(x)| & \leq \int_{(-\delta / \epsilon , \delta / \epsilon)} j(t)|f(x-\epsilon t)-f(x)| \, dt + \int_{\mathbb{R} \smallsetminus (-\delta / \epsilon , \delta / \epsilon)} j(t)|f(x-\epsilon t)-f(x)| \, dt \\ & < \eta \int_{(-\delta / \epsilon , \delta / \epsilon)} j \\ & < \eta . \end{align} Since $x \in \mathbb{R}$ was arbitrary, the above argument shows that $f_\epsilon \rightarrow f$ uniformly.
For the second assertion, since $f_\epsilon$ has compact support in a fixed compact set (you can see this from the definition of $f_\epsilon$), so does $f_\epsilon-f$.
Set $[-K,K] \subset \mathbb{R}$, $K > 0$, a compact set that contains the support of $f_\epsilon-f$. For any fixed but arbitrary $\eta > 0$ there is some $\eta_1 > 0$ such that $$\sup_{x \in \mathbb{R}} |f_\epsilon(x)-f(x)| < \sqrt{\frac{\eta}{2K}} \quad \text{ for all } \epsilon < \eta_1,$$ by the uniform convergence previously shown. Then we see that $$\int_\mathbb{R} |f_\epsilon(x)-f(x)|^2 \, dx = \int_{[-K,K]} |f_\epsilon(x)-f(x)|^2 \, dx < \frac{\eta}{2K} \int_{[-K,K]} dx = \eta ,$$ from which we finally conclude that $f_\epsilon \rightarrow f$ in $L^2(\mathbb{R}).$