Explanation for proof of the continuity of the function in $t$ defined as the integral of a continuous function of two variables

Statement: Let $f:[a,b]\times\mathbb{R} \to \mathbb{R}, (x,t) \mapsto f(x,t)\in\mathbb{R}$ be a continuous function. Then$$F(t) := \int_a^b f(x,t)dx$$is continuous.

I understand the proof for this up until the last part where continuity of $F$ is shown.

Let $t_0\in\mathbb{R}$ and $r>0$ and let $\epsilon>0$. Since $f$ is continuous, it is uniformly continuous in $X=[a,b] \times [t_0-r,t_0+r]$. Therefore we have that $\exists \delta = \delta(t_0,\epsilon)$ s.t.

$$\|(x,t)-(y,s)\|_2<\delta \implies |f(x,t)-f(y,s)|<\epsilon$$for $(x,t),(y,s) \in X$. Now let $h\in\mathbb{R}$ s.t. $|h|<\min\{r,\delta\}$. Then\begin{equation}|F(t_0+h,t_0) - F(t_0)| = \left|\int_a^b(f(x,t_0+h)-f(x,t_0))dx\right| \le (b-a)\epsilon \tag 1\end{equation}

by using uniform continuity of $f$ in the integrand and noting that the arguments are less than $\delta$ apart.

Now, the part I am confused on is how this implies continuity of $F$ at $t_0$? I am slightly rusty on these topics.

I believe we can show $\lim_{h\to0}F(t_0+h)=F(t_0)$ and this is continuity of $F$ at $t_0$. How do I show this limit precisely from (1)? Intuitively my thoughts are that as $\epsilon$ is arbitrarily small, $h$ also gets smaller as $\epsilon$ is small.