Answer
$W$ is not a subspace of $R^2$.
Work Step by Step
Let $W$ be a subset of $V$ such that
$$W=\{(x, y) : y=a x, a \text { is an integer }\}, \quad V=R^{2}.$$
Assume that $u=(x,ax), v=(y,by)\in W$ where $a,b \in Z$ and $c\in R$. Now, we have
(a) $W$ contains the zero vector $(0,0)$.
(b) $u+v=(x,ax)+(y,by)=(x+y,ax+by)$.
Since $ax+by\neq t (x+y)$, $t\in Z$, then $u+v\not \in W$. Hence, $W$ is not a subspace of $R^2$.
