Question: Let $U, V \subset \mathbb{R}^d$ be compact, $K \in C(U \times V)$ and $T_K: C(V) \rightarrow C(U)$ given by $$ T_K(u)(x)=\int_V K(x, y) u(y) d y . $$
Prove that $T\left(B_1(0)\right)$ is precompact in $C(U)$.
My proof:
We need to prove that $T_K\left(B_1(0)\right)$ is precompact in $C(U)$. To do this, we'll show that the closure of $T_K\left(B_1(0)\right)$ is compact in $C(U)$. Since $U$ and $V$ are compact and $K$ is continuous on $U \times V, K$ is uniformly continuous and bounded on $U \times V$. Let's denote $M=\sup \{|K(x, y)|:(x, y) \in U \times V\}$.
Given any $\epsilon>0$, there exists a $\delta>0$ such that for all $x, x^{\prime} \in U$ and $y \in V$, if $\| x-$ $x^{\prime} \|<\delta$, then $\left|K(x, y)-K\left(x^{\prime}, y\right)\right|<\frac{\epsilon}{\mu(V)}$, where $\mu(V)$ is the measure of $V$.
Since $U$ is compact, we can find a finite $\delta$-net for $U$, say $\left\{x_1, x_2, \ldots, x_n\right\}$. This means for every $x \in U$, there exists an $x_i$ such that $\left\|x-x_i\right\|<\delta$.
Now, define functions $\phi_i \in C(U)$ by $\phi_i(x)=\int_V K\left(x_i, y\right) u(y) d y$ for each $i$. These functions are well-defined because the set $\left\{u(y):\|u\|_{C(V)} \leq 1\right\}$ is bounded.
For any $u \in B_1(0)$ and $x \in U$, we can find an $x_i$ such that $\left\|x-x_i\right\|<\delta$. Then, the difference between $T_K(u)(x)$ and $\phi_i(x)$ can be estimated as: $$ \begin{aligned} & \left|T_K(u)(x)-\phi_i(x)\right|=\left|\int_V\left(K(x, y)-K\left(x_i, y\right)\right) u(y) d y\right| \leq \int_V \mid K(x, y)- \\ & K\left(x_i, y\right) \mid d y<\epsilon \end{aligned} $$
This implies that every $T_K(u)$ for $u \in B_1(0)$ is within $\epsilon$ of some $\phi_i$.
Since $\left\{\phi_1, \phi_2, \ldots, \phi_n\right\}$ is a finite set and every $T_K(u)$ for $u \in B_1(0)$ is approximated within $\epsilon$ by this finite set, it means that $T_K\left(B_1(0)\right)$ is totally bounded in $C(U)$.
In conclusion, $T_K\left(B_1(0)\right)$ being totally bounded in the complete space $C(U)$ implies that its closure is compact, and hence $T_K\left(\dot{B}_1(0)\right)$ is precompact in $C(U)$.
-- Is my proof correct? It will be very kind of you if you can proofread it! :)