The part "if $f$ may be extended" is needed, because not every continuous function can be extended onto the closure of its domain: for example, $1/x$ defined on $\mathbb{R}_{+}$ cannot be continuously extended onto $\overline{\mathbb{R}}_{+}$ . Another example is $f(x)=\begin{cases} 1 & \mbox{ if }x>0\\ -1 & \mbox{ if }x<0 \end{cases}$ defined on $A=\mathbb{R}-\{0\}$ which cannot be continuously extended onto $\overline{A}=\mathbb{R}$ .

The part “let $Y$ be Hausdorff” is needed for the uniqueness of a possible extension. Indeed, if $\Delta$ is not closed, then there are some points $(a,b)\in\overline{\Delta}$ such that $a\neq b$ . But then $C'=\{x\in\overline{A}|(g(x),h(x))\in\overline{\Delta}\}$ is closed and contains $A$ , and, hence, $C'=\overline{A}$ . In particular, $\overline{A}$ may contain some point $x$ such that $g(x)=a\neq h(x)=b$ . The easiest example is the two element set $Y=\{a,b\}$ with the topology $\{\emptyset,\{a\},Y\}$ . $f:X\rightarrow Y$ is continuous iff $f^{-1}(\{a\})$ is open in $X$ . Let $f:\mathbb{R}-\{0\}\rightarrow Y$ be such that $f(x)=a$ for $x\neq0$ . Then regardless of whether we define $g(0)=a$ or $g(0)=b$ , the resulting function is going to be a continuous function from $\mathbb{R}$ to $Y$ .

As an alternative proof, one can use Exercises 5, 6 and 7 of Chapter 3 Supplementary Exercises: Nets. Namely, take $x\in\overline{A}$ , use Exercise 6 to argue that there are nets in $A$ converging to $x$ , use Exercise 7 to argue that then the value at $x$ is determined by the values at the points of the net, and finally use Exercise 5 to conclude the argument.