Proof $g \circ f: D \to\mathbb{R}^k$ with $(g \circ f) (x) := g(f(x))$ continuous

Question

Let $D \subset \mathbb{R}^n$ and $E \subset \mathbb{R}^m$. Let also $f\colon D\to E$ and $g\colon E\to\mathbb{R}^k$ be images.

How can one prove that if $f$ is continuous at $a\in D$ and $g$ is continuous at $b := f(a) \in E$, then the image $g \circ f\colon D\to\mathbb{R}^k$ with $(g \circ f) (x) := g(f(x))$ is continuous at $a$?

Can I just do the following? (edited)

$$\lim_{n\rightarrow \infty} g(f(x_n))=g(\lim_{n\rightarrow \infty} f(x_n)) = g(f(\lim_{n\rightarrow \infty} x_n))$$

Answer 1

https://www.math.ucdavis.edu/~hunter/m125a/intro_analysis_ch3.pdf
Looks like Theorem 3.18, page 26. I'd offer the solution I used when I had this question, but this was much faster.