Let ${ (X, d) }$ be a metric space. There is a distance preserving embedding ${ X \hookrightarrow V }$ into a normed space ${ V .}$

Consider ${ \mathcal{B}(X, \mathbb{R}) },$ the space of bounded functions on set ${ X },$ with sup norm. Fix ${ a \in X }.$
Now the map ${ X \hookrightarrow \mathcal{B}(X, \mathbb{R}) }$ sending ${ x \mapsto \varphi(x) := ( f _x - f _a ) },$ where ${ f _p (q) := d(p,q) },$ will work.

Well-defined: Let ${ x \in X }.$ Now ${ f _x - f _a \in \mathcal{B}(X, \mathbb{R}) },$ because ${ \vert f _x (t) - f _a (t) \vert }$ ${ = \vert d(x,t) - d(a,t) \vert }$ ${ \leq d(x,a) }.$ Injective: Say ${ f _{x} - f _a = f _{y} - f _{a} }.$ Now ${ d(x,t) = d(y,t) }$ for all ${ t \in X },$ so ${ d(x,y) = 0 ,}$ ie ${ x=y }.$
Distance-Preserving: Let ${ x, y \in X }.$ We need ${ d(x,y) = \lVert \varphi (x) - \varphi(y) \rVert _{\infty} },$ ie ${ d(x,y) = \sup _{t \in X} \vert d(x,t) - d(y,t) \vert }.$ As ${ \vert d(x,t) - d(y,t) \vert \leq d(x,y) }$ and equality holds when ${ t = y },$ done.

This is called Kuratowski embedding.

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Each ${ \varphi(x) }$ is a Lipschitz function on ${ (X,d) },$ as ${ \vert \varphi(x) (t _1) - \varphi(x) (t _2) \vert }$ ${ = \vert (f _x - f _a)(t _1) - (f _x - f _a)(t _2) \vert }$ ${ \leq \vert d(x, t _1) - d(x,t _2) \vert + \vert d(a, t _1) - d(a,t _2) \vert }$ ${ \leq 2 d(t _1, t _2) }.$

So ${ \varphi }$ infact gives a distance preserving embedding ${ X \hookrightarrow \mathcal{BL}(X, \mathbb{R}) },$ into the space of bounded lipschitz functions on ${ X }$ with sup norm.