Goal: I want to prove $f$ in continuous on $[a,b]$ implies $F(x)=\sup f([a,x])$ is continuous on $[a,b]$.
Proof:
Suppose $f$ in continuous on $[a,b]$.
Let $c \in (a,b)$. Let $\epsilon>0$.
By the continuity of $f$ at $c$, there exists $\delta>0$, such that $-\epsilon/2 +f(c)<f(x)<\epsilon/2 +f(c)$ for all $x \in (c-\delta, c+\delta)$
so $f(x)<\epsilon/2 +f(c)$ for all $x \in [c , c+ \frac{\delta}{2}]$
hence, $\sup f([c , c+ \frac{\delta}{2}]) \le\epsilon/2+f(c) \le \epsilon/2 +\sup f([a,c])$.
so if $y \in f([c , c+ \frac{\delta}{2}]) \cup f([a,c])$, then $y \le \epsilon/2 +\sup f([a,c])$
hence, $\sup f([a ,c+ \frac{\delta}{2}]) \le \epsilon/2 +\sup f([a,c])< \epsilon +\sup f([a,c])$.
by the monotone property of suprema we have $\sup f([a ,x])< \epsilon +\sup f([a,c])$ for all $x \in (c- \frac{\delta}{2},c+ \frac{\delta}{2})$
moreover,
suppose $\sup f([a,c])-\sup f([a,c-\frac{\delta}{2}])\ge \epsilon$
in this case we must have: $\sup f([c-\frac{\delta}{2},c])-\sup f([a,c-\frac{\delta}{2}])\ge \epsilon$ because otherwise $0 \ge \epsilon >0$
then, $\sup f([c-\frac{\delta}{2},c])\ge \epsilon +\sup f([a,c-\frac{\delta}{2}]) \ge \epsilon + f(c-\frac{\delta}{2})$
by the extreme value theorem, there exists $t \in [c-\frac{\delta}{2},c]$ s.t.
$|f(t)-f(c-\frac{\delta}{2})| \ge \epsilon$
but $|f(t)-f(c)|<\epsilon/2$ and $|f(c-\frac{\delta}{2})-f(c)|<\epsilon/2$
so, $|f(t)-f(c-\frac{\delta}{2})|\le|f(t)-f(c)|+|f(c-\frac{\delta}{2})-f(c)|<\epsilon/2+\epsilon/2=\epsilon$, which is a contradiction.
so we must have $\sup f([a,c])-\sup f([a,c-\frac{\delta}{2}])< \epsilon$.
by the monotone property of suprema $\sup f([a,c])-\epsilon< \sup f([a,c-\frac{\delta}{2}])\le \sup f([a,x])$ for all $x \in (c- \frac{\delta}{2},c+ \frac{\delta}{2})$
in conclusion, $|x-c|<\delta/2 \implies \sup f([a,c])-\epsilon< \sup f([a,x])< +\sup f([a,c])+\epsilon \implies |\sup f([a,x])-\sup f([a,c])|<\epsilon$
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I would really appreciate it if community members could check the details of my proof. Are there any mistakes? Is there anything weird about it? What are your general thoughts? I have tried to include as much justification as I thought reasonable, but is there anything you find lacking? Note that I have only attempted to prove continuity on $(a,b)$ and will update to deal with the endpoints later.
Your proof is correct, but somewhat laborious. Here is my suggestion.
We have to show that for each $c \in [a,b]$ and each $\epsilon > 0$ there exists $\delta > 0$ such that $\lvert x - c \rvert < \delta$ implies $\lvert F(x) - F(c) \rvert < \epsilon$, i.e. $F(c) - \epsilon < F(x) < F(c) + \epsilon$. Since $F$ is monotonically increasing, this is equivalent to the existence of $\delta > 0$ such that
Since $f$ is continuous, there exists $\delta > 0$ such that $\lvert \xi - c \rvert < \delta$ implies $\lvert f(\xi) - f(c) \rvert < \epsilon/3$. This $\delta$ will do for 1. and 2.
1.:
For $c \le x < c + \delta$ we have
$$F(x) = \sup f([a,x]) = \max(\sup f([a,c]), \sup f([c,x]) = \max(F(c),\sup f([c,x]).$$ If $\xi \in [c,x]$ we have $\lvert \xi - c \rvert < \delta$, thus $f(\xi) < f(c) + \epsilon/3$ which implies $\sup f([c,x]) \le f(c) + \epsilon/3 < f(c) + \epsilon \le F(c) + \epsilon$. Therefore $\max(F(c),\sup f([c,x]) < F(c) + \epsilon$.
2.:
For $c - \delta < x \le c$ we have $$F(c) = \sup f([a,c]) = \max(\sup f([a,x]), \sup f([x,c]) = \max(F(x),\sup f([x,c]).$$ If $\xi \in [x,c]$ we have $\lvert \xi - c \rvert \le \lvert x- c \rvert < \delta$, thus $$f(\xi) = f(x) + f(\xi) - f(x) \le f(x) + \lvert f(\xi) - f(x) \rvert \le f(x) + \lvert f(\xi) - f(c) \rvert + \lvert f(c) - f(x) \rvert \\ < f(x) + \epsilon/3 + \epsilon/3$$ which implies $\sup f([x,c]) \le f(x) + 2\epsilon/3 < f(x) + \epsilon \le F(x) + \epsilon$, hence $\max(F(x),\sup f([x,c]) < F(x) + \epsilon$.
Note that these arguments also cover the cases $c = a$ and $c = b$.