Let $H$ be a subspace of $V$.
For $c \in V$, define $E(c) = \{c + h\,|\,h\in H\}$
Show that if $a_1, a_2, b_1, b_2 \in V$ such that $E(a_1) = E(b_1)$ and $E(a_2) = E(b_2)$, then $E(a_1 + a_2) = E(b_1 + b_2)$
then show that for any $x\in\mathbb{R}, E(xa_1) = E(xb_1)$.
Since $E(a_1) = E(b_1)$, we have that $a_1$ and $b_1$ differ from an element of $H$, and the same can be said for $a_2$ and $b_2$. So we have that $a_1 = b_1 + h_1$ and $a_2 = b_2 + h_2$.
That way, $E(a_1+a_2) = \{a_1+a_2+h\ |\ h\in H\} = \{a_1+a_2+h_1+h_2+h\ |\ h\in H\} = E(b_1+b_2)$. The second equality holds because any $h'\in H$ can be written in the form $h_1+h_2+h$ for some $h\in H$ (namely $h' = h_1 + h_2 + (h'-h_1 - h_2)$).
The second part holds in a similar way. If $x\neq 0$, it has an inverse and so any $h\in H$ can be written as $h = xh'$ for some $h'\in H$.