Prove that at all points in its domain, a monotone function mapping an open set to $\mathbb R$ is either continuous or has a jump discontinuity. (A jump discontinuity is a point where $\lim^- \neq \lim^+$, as opposed to removable discontinuities or essential discontinuities.)
Note: Proofs are available. This question is to verify, critique, and improve my proof and its exposition.
Lemma: Let $A$ be an open subset of $\mathbb R$ and let $f: A \to \mathbb R$ be an increasing function. Then for all $c \in A$, $\lim_{x \to c^-} f(x)$ and $\lim_{x \to c^+} f(x)$ are defined, and $\lim_{x \to c^-} f(x) \leq f(c) \leq \lim_{x \to c^+} f(x)$.
Proof: Let $\ell = \sup \{f(x) : x \in A, x < c\}$. This set is non-empty and bounded by $f(c)$, and hence has a supremum. For any $\varepsilon > 0$, there exists an $a \in A, a < c$ where $\ell - \varepsilon < f(a) \leq \ell$, for if there were not, then $\ell - \varepsilon$ would be an upper bound. Since $f$ is increasing, for all $x$ such that $a < x < c$ we have $\ell - \varepsilon < f(a) \leq f(x) \leq \ell$, so $\ell = \lim_{x \to c^-} f(x)$. For a similar reason, $f(c) \geq \ell$.
Main Proof: For $x \in A$, we have $\ell := \lim_{x \to c^-} f(x)$ by the lemma and $m := \lim_{x \to c^+} f(x)$ by a similar argument. Suppose that $\ell = m$, and recall that if $\lim^- = \lim^+$, then the limit is defined and equal to $\lim^-$. Then since $\ell \leq f(c) \leq m$, $f(x) = \ell$, and the function is equal to its limit, and therefore continuous, at $x$.
If, alternatively, $\ell \neq m$, we have by definition a jump discontinuity at $x$.
Discussion: Is this proof correct and rigorous? Well-written? How can it be improved?