Topological vector space with discrete topology is the zero space

Let $\Bbb K\in\{\Bbb R,\Bbb C\}.$ Since the scalar multiplication $\Bbb K\times V\to V$ is required to be continuous and $V$ has the discrete topology, the following must hold:

For each $k\in\Bbb K$ and each $v\in V$ there is an $\epsilon>0$ such that for each $l\in B_\epsilon(k)$ we have $lv=kv$.

But as there is always an $l\ne k$ in the ball around $k,$ we get $(l-k)v=0.$ Since $l-k$ is not zero, $v$ must be the zero vector.
This also shows that if $V\ne\{0\},$ then the underlying field has to be discrete.

Another argument: A topological vector space over $\Bbb{R}$ or $\Bbb{C}$ is path-connected, since the map $t\mapsto(1-t)x+ty$ is continuous. The only connected discrete space consists of one point.