In this problem, I need to prove $\sup_{{a}\le{t}\le{x}}f(t)$ is continuous, and I denote this function by $M(x)$. It is obvious that the function $M(x)$ is a non-decreasing function，so I can conclude that $M(x-0) \le M(x)$. Then I just need to prove that the inequality is still true after the inverse. I want to prove that $M(x-0)+\varepsilon\gt{M(x)}$ and because of the arbitary of $\varepsilon$, I conclude that $M(x-0)\geq{M(x)}$. Then $M(x)$ is left-continuous, and the right side case is similar.
I have tried to prove that $M(x-0)+\varepsilon\gt{M(x)}$ is true because of the continuity of $f(x)$, then the following holds. $\forall\varepsilon\gt{0},\exists\delta\gt0$ such that $f(x)-\varepsilon\lt{f(y)}\lt{f(x)}+\varepsilon$ when $y\in(x-\delta,x]$. Then I divide the interval $[a,x]$ into two subintervals which are $[a,y)$ and $[y,x]$, where $x-\delta\lt{y}\lt{x}$. Then $f(y)+2\varepsilon\gt{f(z)}$ for any $z$ in $(x-\delta,x]$, so I can conclude that $$f(y)+2\varepsilon\geq\sup_{x-\delta\lt{z}\leq{x}}{f(z)}$$ then the inequality $$\sup_{{x-\delta}\lt{z}\leq{y}}f(z)+2\varepsilon\geq\sup_{x-\delta\lt{z}\leq{x}}{f(z)}$$ holds. Hence $$\sup_{a\leq{z}\leq{y}}f(z)+2\varepsilon\geq\sup_{a\leq{z}\leq{x}}{f(z)}$$holds. Let $y\to{x}$ from the left side, then:$$M(x-0)+2\varepsilon\geq{M(x)}$$ My problem is that I am not sure my proof is valid and appropriate, and is the right side case really similar to the left side case which has been proved? And I'll be thankful if someone provides a simpler way to tackle this problem.
An alternative approach is by leveraging the uniform continuity of $f$ on the interval $[\alpha, \beta]$. To this end, let $\epsilon >0$. Since $f$ is uniformly continuous on $[\alpha, \beta]$ there would exists $\delta >0$ such that $$\tag{1} \left|f(x)-f(y)\right| < \epsilon/2,$$
for every $x,y\in [\alpha, \beta]$ and $|x-y|< \delta$. Now you can easily prove for example that
$$\tag{*} \lim_{x\to x_0^+}M(x) = M(x_0).$$
Indeed, let $\epsilon >0$ and let $x\in (x_0, x_0+\delta)$. Then, as you mentioned, since $M(x)$ is increasing one has that $M(x_0)\leq M(x)$. Now since
$$M(x) = \sup_{y\in [\alpha, x]}f(y)$$
there would exists $y_x \in [\alpha, x]$ such that
$$\tag{2} M(x)-\frac{\epsilon}{2} < f(y_x).$$
Now consider the following two cases:
Case 1: $\quad y_x \in [\alpha, x_0]$
In this case one has that $f(y_x) \leq M(x_0)$, and hence $M(x)-\frac{\epsilon}{2} < f(y_x) \leq M(x_0)$. Therefore, since
$$M(x) -\frac{\epsilon}{2}< f(y_x) \leq M(x_0)\leq M(x) < M(x)+\frac{\epsilon}{2},$$
we conclude that $|M(x) - M(x_0)|< \frac{\epsilon}{2}<\epsilon.$
Case 2: $\quad \quad y_x\in [x_0, x]$
In this case we leverage the uniform continuity. To this end, by the uniform continuity and since $|y_x - x_0|<\delta$ one has by $(1)$
$$\tag{3} |f(y_x) - f(x_0)|< \frac{\epsilon}{2}\implies f(y_x) < f(x_0)+\frac{\epsilon}{2}.$$
But now $x_0\in [\alpha, x_0]$, and hence $f(x_0)\leq M(x_0)$. Therefore, by $(3)$ we obtain $f(y_x)<M(x_0) +\frac{\epsilon}{2}.$ Therefore, by $(2)$ we get
$$M(x) - \frac{\epsilon}{2}<M(x_0)+\frac{\epsilon}{2}\implies M(x)-\epsilon<M(x_0)$$.
But now, since $M(x_0)\leq M(x)< M(x)+\frac{\epsilon}{2}$ we again conclude that $|M(x)-M(x_0)|< \epsilon.$
Combining Case 1 & Case 2 we conclude that for each $x\in (x_0, x_0+\delta)$ we have that $|M(x)-M(x_0)|<\epsilon.$ This proves $(*)$. Now, one treats in similar fashion the fact that $\lim_{x\to x_0^-}M(x) = M(x_0).$