First lets do a change of variables to get rid of the x in the exponent. Let $y=xu$. Then we have
$F(x)=\int\limits _{-\infty}^{\infty}f(y)\frac{1}{x\sqrt{2\pi}}\mathrm{e}^{-y^{2}/2x^{2}}\, dy=\int\limits _{-\infty}^{\infty}f(xu)\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2} du.$
Next, what is the definition of continuity? Given $\epsilon>0$, we need to show that for fixed $x$
$\biggr|F(x+\delta)-F(x)\biggr|=\biggr|\int\limits _{-\infty}^{\infty}\left(f((x+\delta)u)-f(xu)\right)\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}\, du\biggr|<\epsilon.\ \ \ \ \ \ (1)$
But if $f$ is bounded, say $|f|\leq M$, then the integral on the right hand side is uniformly bounded by $2M\int\limits _{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2} du=2M$ for all $\delta$. Hence when taking the limit as $\delta\rightarrow 0$ the dominated convergence theorem applies so we can switch the order of integration and the limit. Consequently since $f$ is continuous the limit on the right hand side is zero, and we have $\lim_{\delta\rightarrow 0} F(x+\delta)-F(x)=0$ so that $F$ is continuous.
Alternative:
We could split up the integral instead of using the dominated convergence theorem. For the given $\epsilon$ we could choose $N$ so large that the integral
$\int_{-\infty}^N\frac{2M}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}du+\int_N^\infty \frac{2M}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}du<\frac{\epsilon}{2}.$ On the interval $[-Nx,Nx]$ $f$ will be uniformely continuous so we can choose $\delta$ so small that $|f((x+\delta)u)-f(x)|\leq \frac{\epsilon}{4N}$. This implies $\biggr|\int\limits _{-N}^{N}\left(f((x+\delta)u)-f(xu)\right)\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-u^{2}/2}\, du\biggr|<\frac{\epsilon}{2}.$ Upon adding these inequalities we obtain equation $(1)$ and the proof is complete.