Let $\phi_{\epsilon}$ be an approximate identity, i.e. it has the following three properties:
(a) $|| \phi_{\epsilon} ||_{L^1(\mathbb{R}^n)} \leq c$ for some constant $c>0$. (b) $\int_{\mathbb{R}^n} \phi_{\epsilon} dx =1$ (c) for $\delta >0$ it holds that $\int_{|x| \geq \delta} |\phi_{\epsilon} (x) | dx \rightarrow 0$ as $\epsilon \rightarrow 0$.
I would now like to show that $\lim_{\epsilon \rightarrow 0} || f \ast \phi_{\epsilon} - f||_{L^{\infty}(\mathbb{R}^n)}=0$ for Schwartz-function $f$.
I know that any Schwartz function is bounded and uniformly continuous.
With that, I can show that (by choosing some $\rho >0$):
\begin{equation} |(f\ast \phi_{\epsilon})(x)-f(x)| \leq \int_{B_\rho(0)} |f(x-y)-f(x)||\phi_\epsilon (y)| dy + \int_{\mathbb{R}^n\backslash B_\rho(0)} |f(x-y)-f(x)| |\phi_\epsilon (y)| dx \end{equation}
\begin{equation} \leq \int_{B_\rho(0)} |\phi_\epsilon(y)| dy \operatorname{sup}_{|y|\leq \rho}|f(x-y)-f(x)|+ \int_{\mathbb{R}^n\backslash B_\rho(0)} |\phi_\epsilon(y)|dy \operatorname{sup}_{|y|\geq \rho}|f(x-y)-f(x)| \end{equation}
By uniform boundedness, we have that the supremum which is multiplied with the first interval tends to $0$ as $\rho \rightarrow 0$. The integral itself is less than or equal to 1 since $\phi_\epsilon$ is an approximate identity. The second intervall converges to $0$ by assumption (c). The other part is bounded by two times supremum of $|f(x)|$.
Hence, i have shown that $|(f\ast \phi_{\epsilon})(x)-f(x)|$ goes to $0$ as $\epsilon \rightarrow 0$.
I realised that what I've just shown implies pointwise convergence. But do I also show convergence in $L^\infty$. If not, what is left to show?
Since you know that any Schwartz function is bounded and uniformly continuous, the two intervals tend to zero FOR ALL $x$, that is, you've already proved that $\|f\ast \phi_\epsilon-f\|_\infty$ goes to $0$ as $\epsilon\to 0$.