Proof verification: Suppose $f: D \rightarrow \mathbb{R}$ is monotone. Prove that if $f(D)$ is an interval, then $f$ is continuous

Question

I am working through a theorem in Advanced Calculus, Fitzpatrick. Ch3, Theorem 3.23

Suppose $f: D \rightarrow \mathbb{R}$ is monotone. Prove that if $f(D)$ is an interval, then $f$ is continuous

I solved it a different way than the book, I am looking for confirmation that my method is valid. If not, where did I go wrong?

My attempt I will just do the case when $f$ is monotone increasing.

Assume $f(D)$ is an interval. Our goal is to prove $$\forall x \in D \quad \{x_n\} \text { converges to } x \rightarrow \{f(x_n)\} \text{ converges to } f(x)$$

Let $x$ be arbitrary element of $D$ and assume there exists a sequence $\{x_n\} \subset D$ that converges to $x$. There exists a subsequence of $\{x_n\}$, call it $\{x_{n_k}\}$, that is monotone increasing. It can be shown that since $\{x_n\}$ converges to $x$, then $\{x_{n_k}\}$ converges to $sup \{x_{n_k}\} = x$.

Since $f$ is monotonic increasing, $\{f(x_{n_k})\}$ is monotonic increasing. Note that this implies $\forall k \in \mathbb{N}, f(x_{n_k}) \leq f(x)$, since $x$ is the supremum of the subsequence. Therefore $\{f(x_{n_k})\}$ is a bounded monotonic sequence, which converges to $sup \{f(x_{n_k})\} = f(x)$.

Therefore we have shown $\{f(x_{n_k})\}$ converges to $f(x)$ for arbitrary $x \in D$.

Answer 1

That proof cannot be correct because you did not use the fact that $f(D)$ is an interval (and the statement is surely wrong without that condition).

It is correct that $f(x_{n_k})$ increasing implies that $f(x_{n_k}) \to \sup \{f(x_{n_k})\}$, but $f(x_{n_k}) \le f(x)$ for all $k$ implies only that the supremum is $\le f(x)$. You have not shown that the limit is equal to $f(x)$.

Another (minor) error is that an arbitrary sequence converging to $x$ does not necessarily have an increasing subsequence.

You can use your approach to show that $f$ has a left and a right limit at $x$. Then assume that these are different, i.e. $$ \lim_{y \to x-} f(x) < \lim_{y \to x+} f(x) \, . $$ Now show that $f$ does not take values between those limits, contradicting the assumption that $f(D)$ is an interval.