The sets $f^{{-1}}\bigl((-\infty,-\frac{1}{3}]\bigr)$ and $f^{{-1}}\bigl([\frac{1}{3},\infty)\bigr)$ are disjoint and closed in $A$. Since $A$ is closed, they are also closed in $X$. Since $X$ is normal, then by Urysohn’s lemma and the fact that $[0,1]$ is homeomorphic to $[-\frac{1}{3},\frac{1}{3}]$, there is a continuous function $g\colon X\to[-\frac{1}{3},\frac{1}{3}]$ such that $g\Bigl(f^{{-1}}\bigl((-\infty,-\frac{1}{3}c]\bigr)\Bigr)=-\frac{1}{3}$ and $g\Bigl(f^{{-1}}\bigl([\frac{1}{3},\infty)\bigr)\Bigr)=\frac{1}{3}$. Thus $\left\lvert g(x)\right\rvert\leq\frac{1}{3}$ for $x\in X$. Now if $-\leq f(x)\leq-\frac{1}{3}$, then $g(x)=-\frac{1}{3}$ and thus $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$. Similarly if $\frac{1}{3}\leq f(x)\leq 1$, then $g(x)=\frac{1}{3}$ and thus $|f(x)-g(x)|\leq\frac{2}{3}$. Finally, for $\left\lvert f(x)\right\rvert\leq\frac{1}{3}$ we have that $\left\lvert g(x)\right\rvert\leq\frac{1}{3}$, and so $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$. Hence $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$ holds for all $x\in A$. ∎

First suppose that for any continuous function on a closed subset there is a continuous extension. Let $C$ and $D$ be disjoint and closed in $X$. Define $f\colon C\cup D\to\mathbb{R}$ by $f(x)=0$ for $x\in C$ and $f(x)=1$ for $x\in D$. Now $f$ is continuous and we can extend it to a continuous function $F\colon X\to\mathbb{R}$. By Urysohn’s lemma, $X$ is normal because $F$ is a continuous function such that $F(x)=0$ for $x\in C$ and $F(x)=1$ for $x\in D$.

Conversely, let $X$ be normal and $A$ be closed in $X$. By the lemma, there is a continuous function $g_{0}\colon X\rightarrow\mathbb{R}$ such that $\left\lvert g_{0}(x)\right\rvert\leq\frac{1}{3}$ for $x\in X$ and $\left\lvert f(x)-g_{0}(x)\right\rvert\leq\frac{2}{3}$ for $x\in A$. Since $(f-g_{0})\colon A\rightarrow\mathbb{R}$ is continuous, the lemma tells us there is a continuous function $g_{1}\colon X\rightarrow\mathbb{R}$ such that $\left\lvert g_{1}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})$ for $x\in X$ and $\left\lvert f(x)-g_{0}(x)-g_{1}(x)\right\rvert\leq\frac{2}{3}(\frac{2}{3})$ for $x\in A$. By repeated application of the lemma we can construct a sequence of continuous functions $g_{0},g_{1},g_{2},\ldots$ such that $\left\lvert g_{n}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})^{n}$ for all $x\in X$, and $\left\lvert f(x)-g_{0}(x)-g_{1}(x)-g_{2}(x)-\cdots\right\rvert\leq(\frac{2}{3})^{n}$ for $x\in A$.

Define $F(x)=\sum_{{n=0}}^{\infty}g_{n}(x)$. Since $\left\lvert g_{n}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})^{n}$ and $\sum_{{n=0}}^{\infty}\frac{1}{3}(\frac{2}{3})^{n}$ converges as a geometric series, then $\sum_{{n=0}}^{\infty}g_{n}(x)$ converges absolutely and uniformly, so $F$ is a continuous function defined everywhere. Moreover $\sum_{{n=0}}^{\infty}\frac{1}{3}(\frac{2}{3})^{n}=1$ implies that $\left\lvert F(x)\right\rvert\leq 1$.

Now for $x\in A$, we have that $\left\lvert f(x)-\sum_{{n=0}}^{k}g_{n}(x)\right\rvert\leq(\frac{2}{3})^{{k+1}}$ and as $k$ goes to infinity, the right side goes to zero and so the sum goes to $F(x)$. Thus $\left\lvert f(x)-F(x)\right\rvert=0$ Therefore $F$ extends $f$.∎