Using different characterizations of compactness, continuity, etc.

I won't go into the philosophical discussions you are stimulating here. Instead I propose the following "direct sequential proof" that an upper semicontinuous function on an interval $[a,b]$ is bounded. It is essentially the same as yours, but avoids talking about nonexisting circumstances.

Consider the auxiliary function $$g(x):=e^{-f(x)}\qquad(a\leq x\leq b)\ ,$$ and define $$\alpha:=\inf_{a\leq x\leq b} g(x)\geq0\ .$$ Then for all $n\geq1$ there is an $x_n\in[a,b]$ with $g(x)<\alpha+{1\over n}$, and there is a subsequence $\bigl(x_{n_k}\bigr)_{k\geq1}$ with $$\lim_{k\to\infty}x_{n_k}=\xi\in[a,b]\ .$$ If $f(\xi)=:M$ then by assumption on $f$ there is a $\delta>0$ such that $$f(x)\leq M+1\qquad \forall\>x\in\>]\xi-\delta,\xi+\delta[\>\cap\>[a,b]\ .$$ It follows that for large enough $k$ we have $g\bigl(x_{n_k}\bigr)\geq e^{-M-1}$, so that $$\alpha=\lim_{k\to\infty}g\bigl(x_{n_k}\bigr)\geq e^{-M-1}>0\ .$$ This implies $$f(x)=\log {1\over g(x)}\leq\log{1\over\alpha}\qquad(a\leq x\leq b)\ .$$