I just finished this problem from N.L. Corothers' real analysis book and believe I have the right idea, but finished the proof with the wrong conclusion. Let $(M,d)$ be a metric space.
Let $D$ be dense in $M$. Show that $M$ is isometric to a subset of $\ell_{\infty}(D)$.
The hint is to use the following theorem: Let $D$ be dense in $M$, let $N$ be complete, and let $f: D \to N$ be uniformly continuous. Then, $f$ extends uniquely to a uniformly continuous map $F: M \to N$, defined on all of $M$. Moreover, if $f$ is an isometry, then so is the extension $F$.
let $f: D \to \ell_{\infty}(D)$ be given by $f(x) := (x,x,x,...)$ a constant sequence of $x$. Now since $f$ is uniformly continuous, $f$ is an isometry, and $\ell_{\infty}(D)$ is complete: by the given theorem $f$ extends uniquely to a uniformly continuous map $F: M \to \ell_{\infty}(D)$. Since $f$ is an isometry, so is $F$. Thus $M$ is isometric to $F(M) \subset \ell_{\infty}(D)$.
Is this correct? Any criticism is welcome.
Choose any point $x_0 \in D$. Define function $f_x(y)$, $x, y \in D$ as $f_x(y) = d(x, y) - d(x_0, y)$. From triangle inequality, $d(x, y) - d(x_0, y) \leq d(x_0, x)$ and thus $f_x$ is bounded, so in $\ell_\infty(D)$.
For $x_1, x_2 \in D$, $$|f_{x_1}(y) - f_{x_2}(y)| =\\ |d(x_1, y) - d(x_0, y) - d(x_2, y) + d(x_0, y)| =\\ |d(x_1, y) - d(x_2, y)| \leq\\ d(x_1, x_2)$$, so $\|f_{x_1} - f_{x_2}\| \leq d(x_1, x_2)$. And $|f_{x_1}(x_1) - f_{x_2}(x_1)| = d(x_1, x_2)$. Combining, we get $\|f_{x_1} - f_{x_2}\| = d(x_1, x_2)$, so $F: D \to \ell_\infty(D)$, $F(x) = f_x$ is isometry. By theorem from hint, we can extend $F$ to isometry $M \to \ell_\infty(D)$.