Let $X_1, \ldots, X_\infty$ be discrete random variables taking values in a finite or countable set $E$. Suppose that for each $x\in E$, $$\mathbf{P}\left(X_n = x\right) \to \mathbf{P}\left(X_\infty = x\right)$$ as $n \to \infty$.
It can be shown [Sections 6.1-6.2 of Coupling, Stationarity, and Regeneration (Probability and Its Applications) by Hermann Thorisson] that this implies the existence of a coupling $\left(X_{1}',X_{2}',\ldots, X_{\infty}'\right)$ of $X_1, X_2, \ldots, X_\infty$ and a finite random integer $K$ such that $$n\ge K \implies X_n'=X_\infty'.$$
Since $X_n'$ has the same distribution as $X_n$, and $X_\infty '$ has the same distribution as $X_\infty '$, wouldn't the equation $X_n'=X_\infty'$ imply that $X_n$ has the same distribution as $X_\infty$? I don't believe so since I can find variables $X_n$ that converge in probability to some limiting distribution $X_\infty$ such that $X_n$ and $X_\infty$ never have the same distribution.
If you were given $\mathbb P(X_n'=X_\infty') = 1$, that would imply $X_n$ equal to $X_\infty$ in distribution. However, that's not the result you're given. You have $$\mathbb P(X_n'=X_\infty') \ge \mathbb P(K \le n) \to 1$$ as $n \to \infty$, which does not imply equality in distribution for any finite $n$.