Show that $(Tu)(x)=\int_{\alpha(x)}^{\beta(x)} u(t)dt$ is Compact linear operator on $C([0,1])$ Show that
\begin{equation}
(Tu)(x)=\int_{\alpha(x)}^{\beta(x)} u(t)dt
\end{equation} 
is Compact linear operator on $C([0,1],R)$ where $\alpha, \beta:[0,1]\rightarrow [0,1]$ are continuous.
My Attempt
$T$ is obviously linear.
Let $M=\max (\alpha(x)-\beta(x))$. Then
$|Tu(x)|\leq ||u||_{\infty}M$ and $||T||=M$.
For compactness, let $B=\{u\in (C[0,1],R):||u||_{\infty}\leq 1\}$. We show $T(B)$ is equicontinuous family so that by Arzela Ascoli it is relatively compact.
\begin{equation}
|Tu(x)-Tu(y)|=\bigg|\int_{\alpha(x)}^{\beta(x)} u(t)dt-\int_{\alpha(y)}^{\beta(y)} u(t)dt\bigg|
\end{equation}
Please how do I subtract this integrals. I'm thinking to make the assumption $\alpha(y)<\beta(y)<\alpha(x)<\beta(x)$.

The integral from $a$ to $b$ can be viewed as $\int_a^bf=\int_{0}^1\chi_{[a,b]}f$. Recall the relation 
$$
|\chi_A-\chi_B| = \chi_{A \Delta B}
$$
where $A\Delta B$ is the symmetric difference of $A$ and $B$. It is also easy to verify that
$$
[a,b] \Delta [c,d] \subset [a,c] \cup[c,a]\cup[b,d]\cup[d,b] 
$$
where $[x,y]=\emptyset$ if $y<x$, so half of the terms of the right hand side would disappear, depending on the order of $a,b,c,d$. By symmetry of the right hand side, we may assume $a\le c$ and $b\le d$.
Using the above, we can compute that 
$$\begin{align}
\left|\int_a^bf-\int_c^df \right| &\le \int_0^1 |\chi_{[a,b]} - \chi_{[c,d]}| \,|f|\\
&= \int_0^1 \chi_{[a,b] \Delta [c,d]} |f|\\
&\le\int_0^1 \chi_{[a,c]} |f| + \int_0^1 \chi_{[b,d]}|f| \\
&\le (|c-a|+|d-b|)\sup_{x\in[0,1]} |f(x)|.
\end{align}$$
Back to your question, we can deduce that
$$\begin{align}
|Tu(x)-Tu(y)| &\le \big(|\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)| \big)\, ||u||_\infty \\
&\le |\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)|
\end{align}$$
for any $u\in B$. By continuity of $\alpha$ and $\beta$, for any given $\varepsilon>0$ we may find $\delta$ such that 
$$
|x-y|<\delta \implies |\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)|<\varepsilon,
$$
which proves what you want. 
