Let $M,N$ be linear subspaces of $V$. Then if $x,y$ are vectors in V, Show that $(x+M)\cap (y+N) \neq \varnothing \iff x-y \in M+N$. Let $M,N$ be linear subspaces of $V$. Then if $x,y$ are vectors in V,
I am trying to show that $(x+M)\cap (y+N) \neq \varnothing \iff x-y \in M+N$.
Forward direction. Suppose $(x+M)\cap (y+N) \neq \varnothing$, then for some 
$m \in M, n \in N$ $x',y' \in V$, cosets  $x'+m = y'+n \rightarrow x'-y' \in N-M = N+M$.
Backwards direction: Suppose that $x-y \in M+N$ for some $x,y, \in V$, then $x-y = m+n$ for some $m \in M, n \in N$. Then $x-m = y+n$ and so $(x+M)\cap (y+N) \neq \varnothing$. But this seems inadequate to show $x+M \cap y+N$ is not equal to the empty set. Any hints appreciated to show what I am doing wrong.

Your proof for the backwards direction is essentially correct. You have directly constructed a common element of $(x + M)$ and $(y + N)$, which is thus in $(x + M) \cap (y + N)$. However, you should state that $m \in M, n \in N$ and not just $m, n \in V$.
