Let $$ B=\{p\in <!--\; \in \; -->[a,b]:f\left(x\right)\text{ is bounded on }[a,p]\}$$, note that $$ a\in <!--\; \in \; -->B$$ 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 $$ e\in <!--\; \in \; -->B$$ 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 $$ m\in <!--\; \in \; -->B$$, this shows that $$ B$$ is an interval of the form $$ [a,x]$$ for $$ x\in <!--\; \in \; -->[a,b]$$.

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

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

Since $$ \mathbb{R}$$ has the least upper bound property then since $$ B$$ is bounded above by $$ b$$, then we know that $$ s:=\text{sup}\left(B\right)\in <!--\; \in \; -->B$$, 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)