Here's what I have been trying to prove
Suppose that $f$ is continuous on $(a,b]$. Then either there is a point $c_1 \in (a,b]$ such that $f(c_1) \le f(x)$ for all $x\in (a,b]$ or there is a point $c_2 \in (a,b]$ such that $f(x) \le f(c_2)$ for all $x\in (a,b]$.
If there is a point $c_1 \in (a,b]$ such that $f(c_1) \le f(x)$ for all $x\in (a,b]$, then we are done. Suppose not, then $f$ is unbounded below. Now, I've been planning to show that $f$ is bounded above and thus show there is a point $c_2 \in (a,b]$ such that $f(c_2)= \sup \{ f(x) : x \in (a,b]\}$.
Here's the key ideas I've got so far while proving this:
Since $f$ is unbounded below, we can find $x_n \in (a,b]$ for each $n\in \mathbb{N}$ such that $f(x_n)<-n$. Since, $(x_n)$ is bounded, by Bolzano Weierstrass Theorem, there is a subsequence $(x_{n_k})$ converging to some point $c\in [a,b]$. Thus it must be the case that $c=a$ for otherwise by the continuity of $f$, $\lim f(x_{n_k}) = f(c)$ but also have that $\lim f(x_{n_k})=-\infty$ a contradiction.
I need some hints here.