(I have already read the explanations given for the suggested similar questions.)
I approximate that I only understand about 40% of this solution. I have been told that this follows from Kuratowski's Embedding Theorem but I can't find much more than a statement of the theorem. I would like references on where I can get the information needed to comprehend this proof or other elementary proofs of the statement.
Let $X$ be a set and $B(X, \mathbb R)$ the set of bounded functions $f:X\to \mathbb R$ with norm $||f||=\sup\{|f(x)|:x\in X\}$.
Claim: Every metric space $(X,d)$ can be embedded isometrically into the Banach Space $E=B(X,\mathbb R)$.
Proof of claim: Assume $X\neq \varnothing$. Fix a point $a_0\in X$ and for every $a\in X$ define a function $f_a:X\to\mathbb R$ by $f_a(x)=d(x,a)-d(x,a_0)$. The $|f_a(x)|\leq d(a,a_0)$ for every $x\in X$ so $f_a$ is bounded. By setting $\rho : X\to E$, $\rho(a)=f_a$ we have the mapping $\rho : X \to E$.
Claim: $\rho$ is an isometry.
Proof: Let $a,b\in X$. As $x \in X$ we have 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 otherhand (this specifically is a big part I dont understand, why are we looking at $f$ at $a$ now instead of $f$ at $x$?) $|f_a(a)-f_b(a)|=|d(a,a)-d(a,a_0)-d(a,b)+d(a,a_o)|=d(a,b) \therefore ||\rho(a)-\rho(b)||=||f_a-f_b||=d(a,b).$
In the proof of the claim, what is shown first is that $\|f_a-f_b\|\leq d(a,b)$. Then it it shown that $d(a,b)=|f_a(a)-f_b(a)|$. Thus we have $$d(a,b)\leq\sup_{x\in X}|f_a(x)-f_b(x)|=\|f_a-f_b\|,$$ so $d(a,b)\leq\|f_a-f_b\|\leq d(a,b)$, and therefore equality holds.