$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

Let $B=\{p\in [a,b]:f\left(x\right)\text{ is bounded on }[a,p]\}$, note that $a\in B$ since $f\left(\left[a,a\right]\right)=\left\{f\left(a\right)\right\}$ therefore $f\left(\left[a,b\right]\right)$ is bounded.

Suppose that $e\in B$ and $e>a$, then $f$ is bounded on $[a,e]$ by the definition of $B$, meaning that for any $a<m<e$, $f$ is bounded on $[a,m]$ so $m\in B$, this shows that $B$ is an interval of the form $[a,x]$ for $x\in [a,b]$.

$f$ is continuous at $a$, then let $t\in ℝ$ be a throwaway value insofar as to obtain some $\delta \in {ℝ}^{>0}$ so that for all $x\in \text{dom}\left(f\right)=[a,b]$, we have $|x-a|<\delta$ implying $|f\left(x\right)-f\left(a\right)|<t$.

This shows us that $f$ is bounded on $[a,a+\delta ]$ using our throwaway value of $t$, therefore $a+\delta \in B$ and as mentioned in the second paragraph, we know $[a,a+\delta ]\subseteq B$, this means that $B$ has at least two values, so we can find another element $y\in B$ such that $a<y$.

Since $ℝ$ has the least upper bound property then since $B$ is bounded above by $b$, then we know that $s:=\text{sup}\left(B\right)\in B$, from the existance of $y$, then by chaining inequalities we get $a<s$ (todo: define set builder notation) (anything amaller has a point greater def of sup)