Proving function space is a vector space I'm trying to solve this question and have made an attempt at a) but I can't seem to get my head around b). 

Given a set $M$ and a field $K$.
a) Show that $\operatorname{Map}(M,K) := \{f: M \rightarrow K\}$ is a K vector space w.r.t. addition 
$ f+g: M \rightarrow K, m \mapsto f(m) + g(m) $ for all $f,g \in \operatorname{Map}(M,K)$
and scalar multiplication:
$ c \cdot f : M \rightarrow K, m \mapsto c \cdot (f(m))$ for all $ c \in K$ and $ f \in \operatorname{Map}(M,K)$
b) Show that the subset $\{f: M \rightarrow K | f(m) \neq 0 $ for finitely many $ m \in M\}$ is a subspace of $\operatorname{Map}(M,K)$

a) As $ \operatorname{Map}(M,K)$ is the set of all functions from M to K, the function that maps to 0 is also included. This is the neutral element w.r.t. addition. The inverse is defined as $-f(x) \quad \forall x \in M$.
Commutativity of addition: $ (f+g)(m) = f(m) + g(m) = (g+f)(m)$
Commutativity of multiplication: $ ((c+d)\cdot f(m) = (c+d)f(m) = cf(m) + df(m) = (cf)(m) + (df)(m)$
 A: You need to prove that when $f,g$ belong to your set then $f-g$ is also an element of the set. $f$ and $g$ are $0$ everywhere except of finitely many points. Can $f-g$ be non zero when both $f,g$ are zero?
A: For a), you've proved the axioms correctly, but remember there are more axioms to check.
Clearly $\{f:M\to K\mid f(m)\neq 0\text{ for finitely many }m\in M\}$ is a subset of $\operatorname{Map}(M,K)$.  All you have to do is show that it is closed under addition and scalar multiplication.  
To do this, it may help to make the following bit of notation:  For $f\in\operatorname{Map}(M,K)$, define $NZ(f)=\{m\in M\mid f(m)\neq0\}=f^{-1}(K\setminus\{0\})$ (I know it's a bad name, but I couldn't think of anything better).  Now for $f,g\in\operatorname{Map}(M,K)$, and $\lambda\in K$, given $NZ(f)$ and $NZ(g)$, what can you say about $NZ(f+g)$ and $NZ(\lambda f)$?
Finally, I believe $\{f:M\to K\mid f(m)\neq 0\text{ for finitely many }m\in M\}$ should actually be $\{f:M\to K\mid f(m)\neq 0\text{ for *at most* finitely many }m\in M\}$.
