I've been trying to construct my own proof of this, but the only reference I've been able to find that I could (somewhat) understand was the following. In particular, I'm confused at the use of $f_{a}(a)$ and $f_{b}(a)$ in proving that $\|f_{a} - f_{b}\| = d(a, b)$. I realize there's an almost identical question, but it's ~4 years old and received one response which didn't help my understanding.
The proof goes as follows:
Let $X$ be a set and $B(X, \mathbb{R})$ the set of bounded functions $f: X \rightarrow \mathbb{R}$ with norm $\|f\| = \sup_{x \in X}\{|f(x)|\}$. Then every metric space $(X, d)$ can be embedded isometrically into the Banach space $E = \text{Bou}(X, \mathbb{R})$.
Proof. One can assume $X \neq \emptyset$. Fix a point $a_{0} \in X$ and for every $a \in X$ define a function $f_{a}: X \rightarrow \mathbb{R}$ by $$ f_{a}(x) = d(x, a) - d(x, a_{0}). $$ Then $|f_{a}(x)| \leq d(a, a_{0})$ for every $x \in X$ so $f_{a}$ is bounded. Set $g: X \rightarrow E$, $g(a) = f_{a}$. Then we have the mapping $g: X \rightarrow E$, and it now remains to prove that $g$ is an isometry.
Let $a, b \in X$. As $x \in X$, it can be shown that $$ |f_{a}(x) - f_{b}(x)| = |d(x, a) - d(x, b)| \leq d(a, b). $$ Therefore $\|f_{a} - f_{b}\| \leq d(a, b)$. On the other hand $$ |f_{a}(a) - f_{b}(a)| = |d(a, a) - d(a, a_{0}) - d(a, b) + d(a, a_{0})| = d(a, b). $$ Therefore $\|g(a) - g(b)\| = \|f_{a} - f_{b}\| = d(a, b)$.
End Proof.
I can understand how the proof is able to show that $|f_{a}(x)|$ is bounded for every $x$. I also understand how they derived the inequality $|f_{a}(x) - f_{b}(x)| \leq d(a, b)$. Where I'm lost is how the use of $|f_{a}(a) - f_{b}(a)| = d(a, b)$, shows that the equality holds (required for an isometric embedding). I would assume the fact that just prior, we established that $|f_{a}(x) - f_{b}(x)| \leq d(a, b)$ challenges the idea that the equality case is meaningful.
To summarise the discussion in the comments, fix $a,b\in X$. By definition,
$$\|g(a) - g(b)\| = \sup_{x\in X}|[g(a) - g(b)](x)| = \sup_{x\in X}|(f_{a} - f_{b})(x)| = \sup_{x\in X}|f_{a}(x) - f_{b}(x)|.$$
So the result follows by showing that
$$\sup_{x\in X}|f_{a}(x) - f_{b}(x)| = d(a,b).$$
Showing that $|f_{a}(x) - f_{b}(x)| \leq d(a,b)$ for all $x\in X$ allows you to conclude that $d(a,b)$ is an upper bound for $\{|f_{a}(x) - f_{b}(x)| : x\in X \}$, from which it follows that
$$\sup_{x\in X}|f_{a}(x) - f_{b}(x)| \leq d(a,b).$$
On the other hand, showing that $|f_{a}(a) - f_{b}(a)| = d(a,b)$ allows you to conclude that
$$d(a,b) = |f_{a}(a) - f_{b}(a)| \leq \sup_{x\in X}|f_{a}(x) - f_{b}(x)|.$$
This obtains the desired equality.