= Solution
Let $\|K\|_\infty=\max_{[0,1]^2}|K|$. For $f,g\in C([0,1])$,
$$
|Tf(x)-Tg(x)|
\leq\int_0^1|K(x,y)|\,|f(y)-g(y)|\,dy
\leq\|K\|_\infty\|f-g\|_\infty.
$$
Taking the supremum over $x$ proves that the <integral operator> $T$ is <Lipschitz continuous>, hence continuous, in the <uniform norm>.
It remains to see that $Tf$ is continuous. Since $K$ is <uniform continuity>[uniformly continuous] on the compact square and $f$ is bounded,
$$
|Tf(x)-Tf(x')|
\leq\|f\|_\infty\int_0^1|K(x,y)-K(x',y)|\,dy\to0
$$
as $x'\to x$. Therefore $T:C([0,1])\to C([0,1])$ is well defined and continuous.
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