Note that your embedding map $T$ actually takes values in the subspace
$\newcommand{\R}{{\mathbb R}}$
$c_{00}(X;\R)$ of finitely supported functions $X\to\R$. If you merely want a norm on this subspace which makes $T$ an embedding, then this *is* possible via the Arens–Eells construction:

R. Arens, J. Eells, *On embedding uniform and topological spaces.*
Pacific J. Math. 6 (1956) no. 3, 397-403.

(Arens and Eells proved a more general result: if you just want the embedding theorem for metric spaces then it is in Weaver's book *Lipschitz spaces* and also in some more recent work of e.g. Godefroy and Kalton. Google should provide links to various downloadable papers/preprints.)

The embedding is usually phrased in terms of sending $x\in X$ to $\delta_x \in c_{00}(X;\R)$, which is just another way of describing your map $T$. Of course the problem is defining the norm! One can either define it as an inf over various representations or a sup when paired with another more familiar Banach space. Let me choose the second way.

Start by fixing a basepoint $x_0\in X$. Given $f\in \R^X$ with $f(x_0)=0$ define its Lipschitz norm to be
$$ \Vert f\Vert_L = \sup_{x,y\in X; x\neq y} \frac{|f(x)-f(y)|}{d(x,y)} \in [0,\infty] .$$
Then, given $c=\sum_{x\in X} c_x \delta_x$ where only finitely many of the $c_x$ are non-zero, define
$$ \Vert c \Vert_{\bf AE} = \sup\left\{ \sum_{x\in X} c_x f(x) \;\colon\; f\in\R^X, \Vert f\Vert_L\leq 1, f(x_0)=0 \right\}$$.

The completion of $c_{00}(X;{\mathbb R})$ with respect to the norm $\Vert\cdot\Vert_{\bf AE}$ is the Arens–Eells space of $X$ (I'm using the terminology and borrowing the definition from Weaver's book.)

Let's check that $x\mapsto\delta_x$ is an isometry. Let $x,y\in X$ with $x\neq y$. If $f(x_0)=0$ and $\Vert f\Vert_L\leq 1$ then pairing $x$ with $\delta_x-\delta_y$ gives $f(x)-f(y)$, which is bounded in modulus by $d(x,y)$ owing to the Lipschitz condition. So $\Vert \delta_x - \delta_y \Vert_{\bf AE} \leq d(x,y)$. On the other hand, consider the function
$$ h(z)=d(z,y)- d(x_0,y) \quad(z\in X).$$
Clearly $h(x_0)=0$, and the triangle inequality for $d$ shows us that $\Vert h\Vert_L\leq 1$. Hence
$$ \Vert \delta_x -\delta_y \Vert_{\bf AE} \geq \vert h(x)-h(y)\vert = d(x,y). $$
Putting these together gives $\Vert \delta_x - \delta_y \Vert_{\bf AE} =d(x,y)$ as required.

For those who like the category-theoretic perspetive: the Arens–Eells space can be viewed as a left adjoint to the functor ${\bf U}: {\sf Ban} \to {\sf Met}_0$ where:

the first category has Banach spaces as objects and bounded linear maps as the morphisms;

the second category has pointed metric spaces as objects, and basepoint-preserving Lipschitz maps as the morphisms;

and given a Banach space $E$, ${\bf U}(E)$ is defined to be the underlying metric space of $E$, with $0_E$ as the basepoint.

Then the Arens–Eells embedding can be regarded as the unit of this adjunction.

In more "down-to-earth" language: given a pointed metric space $(X,x_0)$ let ${\bf AE}(X,x_0)$ be the Arens–Eells space as defined above. Then for any Banach space $E$ and any Lipschitz map $f: X \to E$ satisfying $f(x_0)=0$, there is a unique extension of $f$ to a continuous *linear map* $F: {\bf AE}(X,x_0) \to E$. Thus ${\bf AE}(X,x_0)$ can be viewed as the "free Banach space generated by $(X,x_0)$".

anynatural norm that is well-defined on all of $\mathbb{R}^X$ (in particular, there does not exist a norm that makes every projection continuous). $\endgroup$ – Eric Wofsey Nov 4 '14 at 17:513more comments