I got a problem with the following exercise:
Show that the subset
$\{ f\colon M \rightarrow K\mid f(m) ≠ 0 \text{ for finite number of } m \in M\}$
is a subspace of $\operatorname{Map}(M,K)$
But since $f(m)$ is not $0$, I can not have the null vector, or am I missing something?
For how many $m\in M$ does the constant zero map assume nonzero values? For none of them, so it belongs to the subset.
If you have a map $f\colon M\to K$, define $\operatorname{supp}(f)=\{m\in M\mid f(m)\ne 0\}$. If $V$ is the subset you're given in the exercise, and $\mathbf{0}$ is the map $\mathbf{0}(m)=0$ (for all $m\in M$), then $\operatorname{supp}(\mathbf{0})=\emptyset$ which is finite.
Note that $\mathbf{0}$ is the null vector in $\operatorname{Map}(M,K)$.
For finishing the exercise, suppose $f,g\in V$ and $a,b\in K$; then prove that $$ \operatorname{supp}(af+bg)\subseteq\operatorname{supp}(f)\cup\operatorname{supp}(g) $$ and conclude that $af+bg\in V$.