So you are having trouble showing that $M$ is closed under countable unions. Here's a walkthrough of how I proved it (other solutions are likely possible): (Added: Indeed, there is a much simpler way of doing it which I thought of much later, added at the bottom)
Show that if $A\in M$ and $B\subseteq V$, then $A$, $B$, $A\cup B$, and $A-B$ are all in $M$.
Let $\{A_i\triangle B_i\}_{i=1}^{\infty}$ be a countable family of elements of $M$. Writing the symmetric difference as $X\triangle Y = (X-Y)\cup (Y-X)$, note that $\bigcup_{i=1}^{\infty}(A_i\triangle B_i) = \left(\bigcup_{i=1}^{\infty}(A_i-B_i)\right) \cup \left(\bigcup_{i=1}^{\infty}(B_i-A_i)\right).$
Show that $\bigcup_{i=1}^{\infty}(B_i-A_i) = \mathcal{B}\subseteq V,$ hence $\mathcal{B}\in M$.
Prove that $\left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcup_{i=1}^{\infty} B_i\right) \subseteq \bigcup_{i=1}^{\infty}(A_i-B_i) \subseteq \left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcap_{i=1}^{\infty}B_i\right).$
Show that $\left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcup_{i=1}^{\infty} B_i\right) \in M.$
Think about what kind of elements can lie in $\left(\left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcap_{i=1}^{\infty}B_i\right)\right) - \left(\left(\bigcup_{i=1}^{\infty}A_i\right) - \left(\bigcup_{i=1}^{\infty} B_i\right)\right).$
Conclude that $\bigcup_{i=1}^{\infty}(A_i-B_i) \in M.$
Conclude that $M$ is closed under countable unions.
Added. In fact, much simpler is to note that $\left(\bigcup_{i=1}^{\infty}A_i\right) - V \subseteq \bigcup_{i=1}^{\infty}(A_i\triangle B_i) \subseteq \left(\bigcup_{i=1}^{\infty} A_i\right)\cup V.$ Now, both the smallest and largest of the three sets lie in $M$, and their difference is a subset of $V$; therefore, the middle set will lie in $M$ by point 1.