Consider $f : (a, b) \rightarrow \mathbb R$ for $a, b \in \mathbb R$ where $f$ is monotone, continuous and bounded. Show that $f$ is uniformly continuous.
Consider a set of $k$ evenly spaced points $p_i \in (a, b)$. Since $f$ is continuous on $(a, b)$ and $k$ is finite, $|f(p_i) - f(p_{i-1})| < \epsilon$ for all $i$ for $k$ large enough. Since $f$ is monotone, $|f(p_i) - f(p_{i-1})| \geq |f(x_1) - f(x_2)|$ for all $x_1, x_2 \in [p_{i-1}, p_i]$. Putting the previous two facts together, $f$ is uniformly continuous on $[p_1, p_k]$.
To extend this to $(a, b)$ we use the fact that $f$ is bounded and so $f(a)$ is arbitrarily close to $f(p_1)$ and $f(b)$ is arbitrarily close to $f(p_k)$ for $k$ large enough.
My question
I am not sure that the only place I need the bounded hypothesis is in the last statement. I think continuity should be enough until the last statement. I think we don't need the bounded hypothesis if we instead consider $f$ over compact sets in $(a, b)$.