This is a continuation of my previous question.
For $c \in V$, define $E(c) = \{c + h\,|\,h\in H\}$.
Prove that if $a, b \in V$ and $E(a) \cap E(b) \neq \varnothing$, then $E(a) = E(b)$.
My initial attempt:
Suppose $E(a) \cap E(b) \neq \varnothing$ and let $u \in E(a) \cap E(b)$. Then there are $\alpha, \beta \in H$ such that $u = a+\alpha = b + \beta$. Since $H$ is a subspace of $V$ and $\alpha,\beta \in H\rightarrow (\alpha-\beta) \in H$ and $$a-b=\beta-\alpha\in H$$
Let $a_1 \in E(a)$, $a_1 = v + a$. Observe: \begin{align*} a_1 = v - (a + b) + (a - b) + a= (v - (a - b)) + b \end{align*} Since $v - (a + b) \in H$, it follows that $a_1 \in E(b)$. Now, we show the other implication:
Let $b_1 \in E(b)$, $b_1 = v + b$. Observe: \begin{align*} b_1 = v - (b + a) + (b - a) + b= (v + (b - a)) + a \end{align*} Since $v - (b + a) \in H$, it follows that $b_1 \in E(a)$. Thus, $E(a) = E(b)$ for $E(a)\cap E(b) \neq \varnothing$.