For the necessity, by Portmanteau theorem (theorem 29.1 in Billingsley or theorem 3.2.21 in Dembo's Notes), for any open interval $(a,b)$ which contains no integer, $P(a which implies that $P(a and further that, for any non-integer $c$, $P(X_\infty=c)=0$ since $P(X_\infty=c) \le \lim_{\epsilon \to 0} P(c-\epsilon.
Using the above two facts, for any integer $k$, we have \begin{align} P(X_\infty=k) & = P(k-0.5$P(X_\infty=k) = \lim_n P(X_n=k)$
For sufficiency, you should assume $X_\infty$ is also an integer valued random variable and then check the point-wise convergence of the distribution functions $P(X_n \le x)$at continuous points of $P(X_\infty \le x)$.
If $X_\infty$ is not integer valued, the converse direction doesn't hold. Here is a counter-example. $P(X_n=k)=1/n$ for $k=1,2,..n$ such that for any $k$, $\lim_{n\to \infty} P(X_n=k)=0$. Let $X_\infty \sim N(0,1)$. Then $P(X_\infty=k)=0$. However, $X_n$ do not converge weakly to $N(0,1)$.