$K(x,y)$ continuous on $S=[0,1]\times[0,1]$ and $\phi \in C([0,1])$, define $T\phi$ on $[0,1]$ by $T\phi(x)=\int_0^xK(x,y)\phi(y)dy,~~x\in[0,1].$ Let $K(x,y)$ be continuous on $S=[0,1]\times[0,1]$ and $\phi \in C([0,1])$, define $T\phi$ on $[0,1]$ by 
$$T\phi(x)=\int_0^xK(x,y)\phi(y)dy,~~x\in[0,1].$$
(a) Show that $T\phi \in C([0,1])$
(b) Show that $T$ is continuous.
(C) If $\{\phi_n\}_{n=1}^\infty$ is a bounded sequence in $C([0,1]),$ show that the sequence $\{T\phi_n\}_{n=1}^\infty$ has a convergent subsequence.
My attempt:
(a) By assumption, we have that $K(x,y)$ is uniformly continuous. So,  $\exists~\delta>0$ s.t. $|K(x,y)-K(x',y')|<\frac{\epsilon}{1+\|\phi\|}$ whenever $\sqrt{(x-x')^2+(y-y')^2}<0$.
$|\int_0^xK(x,y)\phi(y)~dy-\int_0^z K(z,y)\phi(y)~dy| \\ \leq |\int_0^z K(x,y)\phi(y)~dy-\int_0^z K(z,y)\phi(y)~dy +\int_z^x K(x,y)\phi(y)~dy| \\ \leq |\int_0^z (K(x,y)-K(z,y))\phi(y)~dy~| + |\int_z^xK(x,y)\phi(y)~dy| \\ 
 \leq \|\phi\|\frac{\epsilon}{1+\|\phi\|}+|\int_z^xK(x,y)\phi(y)~dy|$
How to estimate this term $|\int_z^xK(x,y)\phi(y)~dy|$ ?
(b) $|\int_0^x K(x,y)\phi(y)~dy|\leq \int_0^1 |K(x,y)\phi(y)|~dy \\ \Rightarrow \|T\phi\|=\displaystyle\sup_{x\in[0,1]}|\int_0^x K(x,y)\phi(y)~dy|\leq \|K\|\cdot\|\phi\|=C\|\phi\|~~~(C=\|K\|)$ 
So, $\|T\phi_1-T\phi_2\|=\|T(\phi_1-\phi_2)\| \leq C\|\phi_1-\phi_2\|$ for $\phi_1,\phi_2 \in C([0,1])$. So, $T$ is continuous.
(c) I have no idea about this problem. Maybe I should show that  $\{T\phi_n\}_{n=1}^\infty$ is equicontinuous. I know the theorem that if  $\{T\phi_n\}_{n=1}^\infty$ is pointwise bounded and equicontinuous on $K$, then  $\{T\phi_n\}_{n=1}^\infty$ contains a uniformly convergent subsequece.
 A: For part $(a)$, you can find a bound for $\left|\int_z^xK(x,y)\phi(y)~dy\right|$ using the following: For $x<z$ we have
\begin{align}
\left|\int_z^xK(x,y)\phi(y)~dy\right|&\leq\int_x^z\left|K(x,y)\phi(y)\right|\ dy\\
&\leq|z-x|\cdot\|\phi\|_{C([0,1])}\cdot\sup_{y\in[x,z]}|K(x,y)|\\
&\leq|z-x|\cdot\|\phi\|_{C([0,1])}\cdot\|K\|_{C([0,1]^2)}
\end{align}
The same bound can be obtained for $z<x$ (just swap $x$ and $z$ where appropriate).
For part (c), you're on the right track to think of the Arzela-Ascoli theorem.  Since $\{\phi_n\}$ is a bounded sequence and $T$ is continuous, $\{T\phi_n\}$ is (uniformly) bounded.  To show it is equicontinuous, for $x_1,x_2\in[0,1]$ with $x_1<x_2$ and $n\in\mathbb N$ we have
\begin{align}
|T\phi_n(x_1)-T\phi_n(x_2)|&=\left|\int_0^{x_1}K(x_1,y)\phi_n(y)\ dy-\int_0^{x_2}K(x_2,y)\phi_n(y)\ dy\right|\\
&=\left|\int_0^{x_1}K(x_1,y)\phi_n(y)\ dy-\int_0^{x_1}K(x_2,y)\phi_n(y)\ dy-\int_{x_1}^{x_2}K(x_2,y)\phi_n(y)\ dy\right|\\
&\leq\int_0^{x_1}|K(x_1,y)-K(x_2,y)||\phi_n(y)|\ dy+\int_{x_1}^{x_2}|K(x_2,y)||\phi_n(y)|\ dy\\
&\leq\int_0^1|K(x_1,y)-K(x_2,y)||\phi_n(y)|\ dy+|x_2-x_1|\|\phi_n\|_{C([0,1])}\|K\|_{C([0,1]^2])}\\
&\leq\|\phi_n\|_{C([0,1])}\left(\sup_{y\in[0,1]}|K(x_1,y)-K(x_2,y)|\right)+|x_2-x_1|\|\phi_n\|_{C([0,1])}\|K\|_{C([0,1]^2])}.
\end{align}
Since $\{\phi_n\}$ is bounded, there exists $M>0$ such that $\|\phi_n\|_{C([0,1])}\leq M$ for all $n$.  Thus, we have
$$|T\phi_n(x_1)-T\phi_n(x_2)|\leq M\left[\left(\sup_{y\in[0,1]}|K(x_1,y)-K(x_2,y)|\right)+|x_2-x_1|\cdot\|K\|_{C([0,1]^2])}\right].$$
Now, you have control over $|x_2-x_1|$, and using the continuity of $K$, you obtain a bound for $\sup_{y\in[0,1]}|K(x_1,y)-K(x_2,y)|$.
