Urysohn's lemma

If $A,B$ are disjoint closed subsets of a normal space, there is a continuous function $f:X\to [0,1]$ with $f{|}_A=0$ and $f{|}_B=1$.

Let open sets $U_t$ be indexed by a dense subset of $[0,1]$, with $\overline{U_s}\subseteq U_t$ for $s<t$. Taking the infimum of membership indices, with value one outside their union, gives a continuous function. The strict sublevel sets are unions of $U_t$; strict superlevel sets are unions of complements of $\overline{U_t}$. These identities prove continuity.

Every continuous real-valued function on a closed subspace of a normal space extends continuously to the whole space. If its values lie in a closed interval, the extension can be kept in that interval.

Suppose $Y$ is closed in a normal space $X$ and $|f|<1$ on $Y$ for a continuous real-valued $f$ on $X$. The closed set $Z=\{|f|\ge 1\}$ misses $Y$. Urysohn's lemma supplies $u:X\to [0,1]$ equal to zero on $Y$ and one on $Z$. Then $(1-u)f$ agrees with $f$ on $Y$ and has range in $(-1,1)$. This permits the inverse transform $s\mapsto s/(1-|s|)$ in the proof of unbounded Tietze extension.

A continuous function on a compact subset of a Tychonoff space extends to the whole space. Embed the space into a cube of closed unit intervals. The Tychonoff theorem makes the cube compact Hausdorff and thus normal. The compact subset is closed in that cube, so the Tietze extension theorem extends its bounded function to the cube. Restrict this extension to the embedded original space. Normality of the original space is unnecessary.

Let $C$ be closed in a normal topological space and $g:C\to [-M,M]$ continuous. Separate its closed level sets $g\le -M/3$ and $g\ge M/3$ by Urysohn's lemma to obtain $h:X\to [-M/3,M/3]$ taking the corresponding endpoint values. Checking the outer and middle ranges gives the displayed bounds. Iteration on residuals gives a uniformly convergent series with total supremum norm at most $M$. The uniform limit theorem gives the bounded Tietze extension theorem.

For a real function $g$ in the bounded continuous functions of norm $M$ on a closed subset $Y$ of a metric space, separate its closed level sets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ by $u=(M/3)(d(\cdot ,A)-d(\cdot ,B))/(d(\cdot ,A)+d(\cdot ,B))$. The distance to a set functions make $u$ continuous with the displayed bounds. Empty level sets are handled by the constants $M/3$, $-M/3$ or $0$. Iterating on the residual yields a uniformly convergent continuous extension with no increase in the supremum norm.