Proof that a composition of continuous functions is continuous

Proof

We work directly with the definition of continuity at a point by open intervals.

1. Let $D, E \subseteq \R$, let $f \colon D \to \R$ with $f(D) \subseteq E$, and let $g \colon E \to \R$. Assume that $f$ is continuous at $a \in D$ and $g$ is continuous at $f(a)$.
2. Let $V$ be an open interval containing $g(f(a))$.
3. Continuity of $g$ at $f(a)$ gives an open interval $W$ containing $f(a)$ with $g(y) \in V$ for all $y \in W \cap E$.
4. Continuity of $f$ at $a$, applied to $W$, gives an open interval $U$ containing $a$ with $f(x) \in W$ for all $x \in U \cap D$.
5. Let $x \in U \cap D$. Then $f(x) \in W$ by step 4, and $f(x) \in E$ since $f(D) \subseteq E$. So $f(x) \in W \cap E$.
6. By step 3, $(g \circ f)(x) = g(f(x)) \in V$.
7. So $U$ works for $V$, and $g \circ f$ is continuous at $a$.
8. If $f$ and $g$ are continuous, then step 7 applies at each $a \in D$, since $g$ is continuous at $f(a) \in E$. So $g \circ f$ is continuous.