Show that the discrete metric can not be obtained from $X\neq\{0\}$ If $X \neq \{ 0\}$ is a vector space. How does one go about showing that the discrete metric on $X$ cannot be obtained from any norm on $X$?
I know this is because $0$ does not lie in $X$, but I am having problems. Formalizing a proof for this.
This is also my final question for some time, after this I will reread the answers, and not stop until I can finally understand these strange spaces.

Suppose$(X,d)$ is a vector space with the discrete metric.
If$X$ is a normed space, then for ∀ $x≠y ∈ X$,$α≠0 ∈K, αx≠αy$.So then $\lVert\alpha x-\alpha y\rVert = 1$
However, by metric and norm properties,
$$\lVert\alpha x-\alpha y\rVert =\lVert\alpha(x-y)\rVert = \lvert\alpha\rvert\,\lVert x-y\rVert\le\vert \alpha \vert.1=\alpha,$$ a contradiction!!
