Let $X_m \xrightarrow{}0$ in probability and $x_mV^2 \xrightarrow{}xV^2$ in distribution where $V\sim \mathcal{N}(0,\sigma^2)$ and $x_m$ is a convergent sequence. Let $X_m \le x_mV^2$ for all $m\in \mathbb{N}$ .
I thought its $$\text{a.s convergence} \Rightarrow \text{convergence in probability} \Rightarrow \text{convergence in distribution},$$ so why is $X_m \rightarrow 0$ almost surely?
It doesn't. Assume $x_m$ is positive and take any example of a sequence of non-positive random variables that converges to zero in probability but not almost surely (e.g. take $U$ uniform over $[0,1]$ and for all $m$, put $k$ s.t. $2^k\leq m+1<2^{k+1}$, $l=m+1-2^k$, and $X_m=-1_{2^k.U\in[l,l+1]}$).
Then the condition $X_m \leq x_m V^2$ is true because by construction $X_m$ is non-positive, however it still does not converge to zero almost surely.
Even if you assume that it is rather $|X_m| \leq x_m V^2$, for $x_m=1$ you can still make my example work by taking $X'_m=X_m 1_{|V|>1}$ which will still be diverging almost surely when $|V|>1$.
If you add a second hypothesis that $x_m$ converges to zero, then it becomes trivial because your sequence $X_m$ is bounded by a sequence that converges to zero a.s.