For $f\in L^2(\mathbb{R})$, denote $$s_N(x)=\dfrac{1}{2\pi}\int_{-N}^N\hat{f}(t)e^{ixt}dt.$$
I'd like to prove that the integral converges, and that $s_N$ is continuous.
Since $f\in L^2(\mathbb{R})$, we know that also $\hat{f}\in L^2(\mathbb{R})$. So $$\left|\int_{-N}^N\hat{f}(t)e^{ixt}dt\right|\leq\int_{-N}^N|\hat{f}(t)e^{ixt}|dt=\int_{-N}^N|\hat{f}(t)|dt\leq\left(\int_{-N}^N|\hat{f}(t)|^2dt\right)^{1/2}\left(\int_{-N}^N1dt\right)^{1/2}$$ The right-most term is finite because $\hat{f}\in L^2(\mathbb{R})$.
Now, why is $s_N$ continuous? It looks like an application of the dominated convergence theorem, but how?
Just differentiate under the integral sign (your limits of integration are finite, so life is good), and you will see that $s_N$ is differentiable, hence continuous.