Uniformly continuous in the compact variable What is this theorem/where can I find a proof of the following?

Let $f: K \times X \to Y$ be continuous, with $K$ compact. Using uniform continuity ideas, for $x \in X$ and $\epsilon > 0$, there exists $\delta = \delta_{x, \epsilon} > 0$ such that $\|x - x'\| < \delta$ forces $\|f(t, x) - f(t, x') \| < \epsilon$ for all $t \in K$. That is, we can pick $\delta$ in a manner which is uniform across $K$ for fixed $x \in X$ and $\epsilon > 0$. In other words, $f$ is "uniformly continuous in the compact variable."
Using this, we can conclude that for a continuous map $f: [a, b] \times X \to \mathbb{R}$, the function $X \to \mathbb{R}$ defined by$$x \mapsto \int_a^b f(t, x)\,dt$$is continuous.

 A: We first justify the application to integrals. For $f: [a, b] \times X \to \mathbb{R}$ a continuous map, the function $f_x: [a, b] \to V$ given by $t \mapsto f(t, x)$ is continuous, so the integral $I(x) = \int_a^b f_x = \int_a^b f(t, x)\,dt$ makes sense. Using the estimate$$|I(x') - I(x)| \le \int_a^b |f(t, x) - f(t, x')|\,dt \le \epsilon(b-a)$$of $\|x' - x\| < \delta = \delta_{x, \epsilon}$ with $\delta$ as in the main part of the problem, we deduce the desired continuity of $I$.
Now we turn to the construction of $\delta_{x, \epsilon}$ for continuous maps $f: K \times X \to Y$ with $K$ compact. We fix $x$ and $\epsilon$, and we use $\|\cdot\|$ to denote the norm arising from choices of inner products on the ambient vector spaces containing each of $K$, $X$, and $Y$. For each $t \in K$, we get some $\delta_t > 0$ such that$$\max(\|t - t'\|, \|x - x'\|) < \delta_t \implies \|f(t, x) - f(t', x') \| < \epsilon.$$The opens $U_t = B_{\delta_t/2}(t)$ form an open covering of the compact $K$, so there is some finite sub covering $U_{t_1}, \dots, U_{t_n}$. Let $\delta = \min \delta_{t_j} > 0$. Now pick any $x' \in X$ with $\|x - x'\| < \delta$, and we wish to prove$$\|f(t, x) - f(t, x') \| < \epsilon$$for all $t \in K$. Fix a choices of $t$, so $t \in U_{t_j} = B_{\delta_j}(t_j)$ for some $j$. We have the estimate$$\|f(t, x) - f(t, x')\| \le \|f(t, x) - f(t_j, x)\| + \|f(t_j, x) - f(t, x')\|$$with the second term on the right at most $\epsilon$ since $\max(\|t_j - t\|, \|x - x'\|) < \max(\delta_{t_j}, \delta) \le \delta_{t_j}$. Meanwhile, we have$$\max(\|t - t_j\|, \|x - x\|) = \|t - t_j\| < \delta_{t_j},$$so $\|f(t, x) - f(t_j, x)\| \epsilon$. Putting everything together, when $\|x - x'\| < \delta$, we get$$\|f(t, x) - f(t, x')\| \le 2\epsilon$$for all $t \in K$. This completes the proof.
