Continuity of primitive function

Suppose $$f : [a,b] \times [c,d]\to\Bbb{R}$$ is continuous. Let $$F:[c,d]\to \Bbb{R}$$ and $$F(y)=\int_a^bf(x,y) \space\space dx.$$

Show $$F(y)$$ is continuous.

I try the following : Let $$y,t \in [c,d]$$. Suppose $$F(y)$$ is not continuous $$\implies$$ $$F(y)$$ is not uniformly continuous. We have that $$\forall \delta>0\exists \epsilon >0$$, s.t $$\vert y-t\vert<\delta\implies\vert F(y)-F(t)\vert=\Biggr\vert\int_a^bf(x,y)-\int_a^bf(x,t)\Biggr\vert=0\geq\epsilon.$$

A contradiction. I feel like this is wrong, since I didnt invoke the continuity of $$f$$ at any point. What should I take a look at to understand this problem?

$$|F(y)-F(t)| = \left|\int_a^b(f(x,y) - f(x,t))dx\right|=0$$ doesn't follow from what you assumed, that is, neither the form of $$F$$ nor that it is not uniformly continuous. Thus you can't get the contradiction $$0\ge \epsilon$$.
By definition of continuity of $$F$$, let $$\epsilon >0$$. We need to show there is some $$\delta>0$$ s.t for $$y, t \in [c, d]$$, $$|y-t|<\delta \implies |F(y)-F(t)|<\epsilon$$
Since $$f$$ is continuous on $$[a,b]\times[c,d]$$, it is uniformly continuous there. Hence, in particular, given $$\epsilon >0$$, there is $$\delta >0$$ s.t. for all $$x\in[a,b]$$, $$|(x,y) - (x,t)|=|y-t| <\delta \implies |f(x,y)-f(x,t)|<\epsilon/(b-a)$$
Thus: $$|F(y)-F(t)| = \left|\int_a^b(f(x,y) - f(x,t))dx\right|\\ \leq \int_a^b \left|f(x,y)- f(x,t)\right| dx\\ \leq \epsilon(b-a)/(b-a) = \epsilon$$