Solutions

Suppose that $f$ is increasing on $(a, b)$ and continuous at $a$ and $b$. Let $y \in (a, b)$. For contradiction, suppose $f(a) > f(y)$. This means that $\varepsilon = f(a) - f(y) > 0$ and by definition that there exists $\delta > 0$ such that

$$\forall x, \quad a < x < a + \delta \implies |f(x) - f(a)| < f(a) - f(y).$$

Now, let $\delta = \min(\delta, y - a)$. Then we get

$$\begin{align*} \forall x,\ a < x < a + \delta \le y &\implies |f(x) - f(a)| < f(a) - f(y) \\ &\implies f(y) - f(a) < f(x) - f(a) < f(a) - f(y) \\ &\implies f(y) < f(x) < 2f(a) - f(y), \end{align*}$$

which contradicts the supposition that $f$ is increasing on $(a, b)$. Thus $f$ is increasing on $[a, b)$. A similar proof by contradiction can be supplied to prove that $f$ is increasing over $(a, b]$ and therefore over the entire interval $[a, b]$.