Finding a norm on $ \mathbb{R}^X $ such that the "natural" embedding of a metric space $ X $ in $ \mathbb{R}^X $ becomes an isometry

Question

This question is related to this one: Every metric space can be isometrically embedded in a Banach space, so that it's a linearly independent set

Kuratowski's embedding indeed seems to be the solution to that problem, but I'm wondering if my initial idea had any actual potential or not.

So, the current problem I'm wondering about is the following:

Let $ (X,d) $ be a metric space and consider the function $ T:X \to \mathbb{R}^X$ such that $ T(x)(y) = 1$ if $ y = x $ and $ 0 $ for all other $ y $. Is there a norm on $ \mathbb{R}^X$ such that $ T $ is an isometry? That is, $ ||T(a) - T(b)|| = d(a,b)$ for all $ a,b \in X $.

I'm at a loss to know how to approach this. I didn't come up with any good ideas on how to define a proper norm, and I have absolutely no clue how to begin trying to prove such a norm could not exist. Any ideas?