 Solution

For $x_0>a$, defineThis nonempty set is bounded above by $f(x_0)$, so $l\le f(x_0)$. Given $\epsilon >0$, the supremum property supplies $s<x_0$ with $l-\epsilon <f(s)\le l$. Whenever $s<t<x_0$, monotonicity givesThis proves the left-hand one-sided limit $f^{-}(x_0)=l$.

For $x_0<b$, defineHere $r\ge f(x_0)$. Choose $u>x_0$ with $r\le f(u)<r+\epsilon$, using the infimum property. For $x_0<t<u$, monotonicity gives $r\le f(t)\le f(u)<r+\epsilon$, proving the right-hand one-sided limit $f^{+}(x_0)=r$. Thuswhere the relevant one-sided limits exist, including the indicated single one-sided limit at each endpoint.

At an interior point, continuity implies both one-sided limits equal $f(x_0)$. Conversely, equality $f^{-}(x_0)=f^{+}(x_0)$ and the displayed inequalities force both to equal $f(x_0)$, giving continuity. Therefore