Let $f$ be continuous on $[a, b]$. Define a function $g$ as follows: $g(a)=f(a)$ and, for $x$ in $(a, b]$ $$g(x)=\sup \{f(y): y \text { in }[a, x]\}$$ Prove that $g$ is monotone increasing and continuous on $[a, b]$.
Regarding the monotone increasing, if $g(x) < g(y)$ for some $x,y$ then using the fact that supremum of any set is greater than or equal to the supremum of any subset of it, we will reveal that $g$ is a non-decreasing function. However, I could not come up with how "monotone increasing" can be deduced; moreover, how one could prove the continuity of $g$ without using the epsilon-delta definition: are there any direct features we could apply in this particular problem?
Your deduction for monotonic increment of $g(x)$ is correct as $[a,x_1]\subset[a,x_2]$ for $x_1<x_2$. For continuity, you must prove that $$ |x_2-x_1|<\delta\implies |g(x_2)-g(x_1)|<\epsilon . $$ Arbitrarily let $x_2>x_1$. Hence $$ g(x_2)=\sup_{y\in[a,x_2]}f(y)=\max\left\{\sup_{y\in[a,x_1]}f(y),\sup_{y\in(x_1,x_2]}f(y)\right\} =\max\left\{g(x_1),\sup_{y\in(x_1,x_2]}f(y)\right\}, $$ hence $$ g(x_2)-g(x_1)=\max\left\{0,\sup_{y\in(x_1,x_2]}f(y)-g(x_1)\right\}. $$ The rest of the proof should now be easy due to the continuity of $f(x)$ over $(x_1,x_2]$.