To see that $T$ is well-defined, note that since $[0,1]\times [0,1]$ is compact and $k$ is continuous, $k$ must be uniformly continuous. Thus there exists some $\delta$ such that $k$ varies by less than $1$ over open balls of radius $\delta$. Since $[0,1]\times [0,1]$ is a finite union of open balls of radius $\delta$, this makes $k$ bounded, say by $C$. Clearly a bounded function times an $L^1$ function is $L^1$, hence the integral is well-defined.
To see that $T$ is compact, you need to show that the image of the unit ball in $L^1([0,1])$ has compact closure in $C^0([0,1])$. The unit ball in $L^1([0,1])$ is the set of functions $f\in L^1([0,1])$ such that $\int_{[0,1]} |f(y)|dy<1$. We want to verify that the image is uniformly bounded and equicontinuous. For uniformly bounded, note that $\left|\int_{[0,1]}k(x,y)f(y)dy\right|\leq \int_{[0,1]}|k(x,y)||f(y)|dy\leq C\int_{[0,1]}|f(y)|dy For equicontinuity note that since $k$ is uniformly continuous for any $\epsilon>0$ we have some $\delta>0$ such that $|x-x'|<\delta$ implies $|k(x,y)-k(x',y)|<\epsilon$. Thus if $|x-x'|<\delta$ we have $\begin{align} \left|\int_{[0,1]}k(x,y)f(y)dy-\int_{[0,1]}k(x',y)f(y)dy\right| &= \left|\int_{[0,1]}(k(x,y)-k(x',y))f(y)dy\right|\\ &\leq \int_{[0,1]}|k(x,y)-k(x',y)||f(y)|dy\\ &\leq \int_{[0,1]}\epsilon|f(y)|dy<\epsilon\\ \end{align}$ hence the image of the unit ball is equicontinuous as well. Thus by Arzela-Ascoli it has compact closure, so $T$ is a compact operator.