Define metric space from normed linear space without using the norm

(If this question may expose some fundamental misunderstanding about normed linear spaces and metric spaces, I appreciate any corrections to these misconceptions in lieu of answering the question directly.)

I understand that if we have a normed linear space $$(X, \|\cdot\|)$$, it is also a metric space with the metric $$d(x,y)=\|x-y\|$$. Is it reasonable/possible/sensible to define a metric space from $$X$$ without using the norm as the metric? e.g. could it make sense to define a metric space from the normed linear space by equipping it with the discrete metric instead?

It makes sense to define the discrete metric on any set. But why would you do that on a real vector space $$V$$? Then, for instance, if you take $$v\in V\setminus\{0\}$$, the map$$\begin{array}{ccc}\mathbb R&\longrightarrow&V\\\lambda&\mapsto&\lambda v\end{array}$$will not be continuous.
• I suppose my question should have been, "why do we say that 'a normed linear space is a metric space' with the specific metric $d(x,y)=||x-y||$?" It seems that we're arbitrarily creating a metric on the original vector space (we just happen to be choosing one that's very reasonable). So, if we're taking a normed vector space and assigning an arbitrary metric to it to make it a metric space, what makes it special? E.g. why can't we extend this same argument to conclude that 'any set is a metric space?'
• On any normed vector space you define, as you wrote, $d(x,y)=\lVert x-y\rVert$. How do you do the same thing on an arbitrary set? Oct 2 '19 at 20:08
• The "argument" would be that since we're defining $d(x,y)=||x-y||$ arbitrarily, we can simply pick another metric e.g. $d(x,y) = 1$ if $x=y$ and $0$ otherwise. This is a weird choice to make, but I can't see the formal reason we aren't allowed to do it. Continuing with this reasoning, since we're imposing the discrete metric $d(x,y)$ on the normed linear space, we can similarly impose the discrete metric onto any set and conclude it's a metric space.
• Yes, we can do that. But then, when we are dealing with a vector space, we get a distance that has no relation whatsoever with the vector space structure. And I must say that I think that it is rather strange that you find arbitrary the choice of taking $d(x,y)=\lVert x-y\rVert$. The norm is already there and the map $(x,y)\mapsto\lVert x-y\rVert$ is a distance! Oct 2 '19 at 21:39