I'm looking for a counter-example:
Let $f:[0,1]\times \mathbb R\to\mathbb R$ be continuous in such a way that
$$F(x):=\int_{\mathbb R} f(x,t) dt$$
defines a function $F:[0,1]\to\mathbb R$ (so in particular the integral is finite).
I'm looking for an $f$ such that $F$ is not continuous (it's enough if $F$ is discontinuous in one point, for example $0$).
Here is an example: Set $g(y) = \frac{1}{1+y^2}$ (so $\int_\mathbb{R} g(y) dy = \pi$) and then $$ f(x,t) = \begin{cases} g\left(t - \frac{1}{1-2x}\right) \quad (x < \frac{1}{2}) \\ 0 \quad (x \ge \frac{1}{2} \end{cases} . $$ It's not hard to see that $f$ is continuous. Then $F(x) = \pi$ for $x < \frac{1}{2}$ and $F(x) = 0$ for $x \ge \frac{1}{2}$.
The effect occurs because $f(x,t) \to 0$ for all $t$ as $x \to \frac{1}{2}$, but the integral does not converge (Lebesgue's Theorem cannot be applied).