Proof if $f(x)$ is continuous on $[a,b]$,strictly increasing on $(a,b)$,then $f(x)$ is strictly increasing on $[a,b]$ My attempt:
Because $f(x)$ is continuous at $a$,then $\lim_{x\rightarrow a^+}f(x)=f(a)$
When $x\rightarrow a^+$,$f(x)$ is strictly decreasing,so for any $x \in (a,b)$,  $f(x)>f(a)$. It can also show that $f(x)<f(b)$ in the same way.
Do I prove it in a rigorous way?Thanks in advance!

The idea is very simple. Since $f$ is strictly increasing on $(a, b)$ it follows that if we have $a < x < b$ then we can choose $y, z$ such that $a < y < z < x < b$ and then $f(y) < f(z) < f(x)$. Letting $y \to a^{+}$ we get $f(a) \leq f(z) < f(x)$ and thus $f(a) < f(x)$. Similarly it can be proved that $f(x) < f(b)$. It follows that $f$ is strictly increasing on $[a, b]$.
