Let $f : [a,b] \times [c,d] \to \mathbb{R} $ be a continuous function. Let $F(t,y) = \int_{a}^{t} f(x,y) dx$ be a function. Show that F is continuous.

Question

Let $f : [a,b] \times [c,d] \to \mathbb{R} $ be a continuous function. Let $F(t,y) = \int_{a}^{t} f(x,y) dx$ be a function. Show that F is continuous.

I attempted to prove the above but I am not sure whether this is correct. It goes as follows:

Consider:

For $(t, y), (t_0, y) \in [a,b] \times [c,d]$

\begin{align*} |F(t, y) - F(t_0, y_0)| & = |\int_{a}^{t} f(x,y) - \int_{a}^{t_0} f(x, y_0)\ dx| \\ & = |\int_{a}^{t_0} f(x,y) dx - \int_{a}^{t_0} f(x,y) dx + \int_{a}^{t} f(x,y) - \int_{a}^{t_0} f(x, y_0)\ dx| \\ & \leq |\int_{a}^{t} f(x,y) dx - \int_{a}^{t_0} f(x,y) dx| + |\int_{a}^{t_0} f(x, y) - f(x, y_0) dx | \end{align*}

And we have continuity since we have from FTC that the if f is integrable (which it is since continuity implies integrability) then $F(x) = \int_{a}^{x} f(t) dt$ is continuous. And the second integral is also continuous since f is continuous on [a,b] x [c,d]. Hence it follows that F is continuous.

Does this hold?