= Solution
A <C-embedded subset> $Y\subseteq X$ is one when every <continuous> <real-valued function> on $Y$ extends continuously to $X$. It is a <Cb-embedded subset> when this is required for every bounded <continuous> <real-valued function>, with a bounded extension. These definitions concern the <subspace topology> on $Y$.
Here is the bounded Tietze extension construction from <Urysohn's lemma>. Let $g:Y\to\mathbb R$ be <continuous> with $|g|\le M$. The case $M=0$ has the zero extension. For $M>0$, the <sets> $A=\{g\le-M/3\}$ and $B=\{g\ge M/3\}$ are disjoint and closed in $X$, since $Y$ is closed. <Urysohn's lemma> gives a <continuous> $u_0:X\to[-M/3,M/3]$ taking the endpoint values on $A,B$. On $Y$,
$$
|g-u_0|\le2M/3.
$$
Indeed on $A$ or $B$ this follows from the prescribed endpoint; on their complement both $|g|$ and $|u_0|$ are at most $M/3$.
Repeat on the residual. With $M_n=M(2/3)^n$, choose <continuous> $u_n$ satisfying $\|u_n\|_\infty\le M_n/3$ and reducing the residual on $Y$ to at most $M_{n+1}$. The <series> $F=\sum_{n\ge0}u_n$ converges uniformly by the geometric majorant. Its sum is <continuous>, has $|F|\le\sum M_n/3=M$, and restricts to $g$, since the residual tends uniformly to zero. Thus \b[every bounded <continuous function> extends, with its bound preserved], proving $C^b$-embeddedness without assuming the extension theorem being proved.
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