= Solution
\b[The extension.] <Urysohn's lemma> states that disjoint <closed subsets> $A,B$ of a <normal topological space> can be separated by a <continuous function> $u:X\to[0,1]$ with $u|_A=0$ and $u|_B=1$.
For the bounded <Tietze extension theorem>, first suppose $g:C\to[-M,M]$ is continuous. The subsets $A=\{g\leq-M/3\}$ and $B=\{g\geq M/3\}$ are closed in $X$, because $C$ is closed. <Urysohn's lemma> supplies a <continuous function> $h:X\to[-M/3,M/3]$ with $h=-M/3$ on $A$ and $h=M/3$ on $B$. Directly checking the two outer ranges and the middle range of $g$ gives
$$
\|h\|_\infty\leq M/3,\qquad\|g-h|_C\|_\infty\leq2M/3.
$$
Empty level sets cause no difficulty: take the appropriate constant, or zero if both are empty. This is the <Urysohn two-thirds approximation>.
Apply this successively to the residual on $C$, beginning with $f$ and $M_0=1$. We get $h_n$ with $\|h_n\|_\infty\leq\tfrac13(\tfrac23)^n$ and residual after the first $n$ terms at most $(2/3)^n$. The <Weierstrass M-test> gives <uniform convergence> of $F=\sum_{n\geq0}h_n$. By the <uniform limit theorem>, $F$ is continuous; the residual bound proves $F|_C=f$, and the sum of the norm bounds is one. Thus
$$
\boxed{F:X\to[-1,1]\text{ is continuous and }F|_C=f.}
$$
\b[Closed sets as exact zero sets.] If $X$ is a <perfectly normal space>, write a closed $C$ as $\bigcap_{n\geq0}U_n$ with each $U_n$ open. Apply <Urysohn's lemma> to $C$ and $X\setminus U_n$ to obtain $f_n:X\to[0,1]$ vanishing on $C$ and equal to one on $X\setminus U_n$. The uniformly convergent series
$$
\boxed{f=\sum_{n\geq0}2^{-n-1}f_n\quad\text{satisfies}\quad f^{-1}(0)=C.}
$$
Indeed all summands vanish on $C$; outside $C$, at least one summand is strictly positive. The <uniform limit theorem> proves continuity. This realizes the <closed G-delta set as a zero set> construction.
Conversely, suppose each closed set is the <zero set of a continuous function>. For disjoint closed $C,D$, select $f,g:X\to[0,1]$ with zero sets $C,D$. Their sum never vanishes, and
$$
\boxed{h=\frac{f}{f+g}:X\to[0,1],\qquad h^{-1}(0)=C,\quad h^{-1}(1)=D.}
$$
The quotient is continuous. This also proves normality by inverse images of disjoint endpoint neighborhoods. Each zero set is a <G-delta set>, since $f^{-1}(0)=\bigcap_{n\geq1}f^{-1}([0,1/n))$; thus this property characterizes <perfectly normal spaces> among <T1 spaces>.
\b[Subspaces.] Let $Y\subseteq X$ and let $A,B$ be disjoint <closed subsets> of $Y$. Put $C=\overline A^{\,X}$ and $D=\overline B^{\,X}$. Since $C\cap Y=A$ and $D\cap Y=B$, the intersection $C\cap D$ misses $Y$. Choose <continuous functions> $f,g$ on $X$ whose zero sets are $C,D$. On the <open neighborhood> $W=X\setminus(C\cap D)$ of $Y$, the same quotient $f/(f+g)$ is continuous and is zero on $A$, one on $B$. Its inverse images of $[0,1/3)$ and $(2/3,1]$, restricted to $Y$, are disjoint <open neighborhoods>. Therefore \b[every subspace is a <normal topological space>]: <perfect normality implies hereditary normality>.
\b[A strict counterexample to the converse.] Let $D$ be an uncountable <discrete space>, and let $X=D\cup\{\infty\}$ be its <one-point compactification>. Its neighborhoods of $\infty$ have finite complement in $D$, because compact subsets of a <discrete space> are finite. The space is compact Hausdorff, hence a <T1 space> and normal. A subspace missing $\infty$ is discrete. A subspace containing $\infty$ is closed in $X$, since its complement is a subset of the open discrete part; it is therefore compact Hausdorff and normal. Thus all subspaces are normal.
But $\{\infty\}$ is not a <G-delta set>. A countable intersection of its <open neighborhoods> omits at most countably many elements of $D$, and cannot omit all of uncountable $D$. Hence $X$ is not a <perfectly normal space>. This is the <uncountable discrete one-point compactification counterexample>.
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