Proceed by contradiction. Pick $A$ and $\varepsilon$ with $A > 2\varepsilon$. If the function $f$ is not bounded, then there exists a sequence $x_n \rightarrow \infty$ such that $f(x_n) > A$. Since $\int_{-\infty}^{\infty} f(x) dx < \infty$ there exists a sequence of $h_n \rightarrow 0$ such that $f(x_n + h_n) < A/2$ (the function $f$ cannot remain large for too long). Hence $f(x_n + h_n) - f(x_n) \geq A/2$. But once $n$ is large enough we will have $f(x_n + h_n) - f(x_n) < \varepsilon$ by uniform continuity, because $h_n \rightarrow 0$. Hence $A/2 \leq f(x_n + h_n) - f(x_n) < \varepsilon$, a contradiction with $A > 2 \varepsilon$.