Let $E,O$ $\subset$ $F(R,R)$ denote the sets of even and odd functions respectively. Prove that the $E$ and $O$ are subspaces.
How do you jump “$E$ is nonempty” to “there exists $0\in E$ so that $a(−x)=0= a(x)$”? That doesn't make sense. The set $\{1\}$ is also nonempty, but it doesn't contain the $0$ function. You can simply say that $0$ is an even function; in other words, $0\in E$.
The rest is correct.
Your proof of the fact the set is closed under addtion and scalar multiplication are fine. But I don't quite understand what you are saying about the zero vector. The zero function $f$ is defined by $f(x)=0$ for all $x$. This function is even and this is the zero element of the space.