Say you have a sequence $\{X_n\}$ of mean zero random variables with finite variance. Say you also have a random variable $X$ such that $$X_n\xrightarrow{D}{}X,$$ i.e., the sequence converges in distribution to $X$.
My question is: is it true that $$\limsup_{n\to\infty}Var(X_n)\leq Var(X),$$ where $Var(X)$ is the variance of X?
No. Let $X$ have any distribution with finite variance and define $X_n = X + \xi_n$, where $\xi_n$ is a non-constant random variable independent of $X$ such that $\xi_n \Rightarrow 0$, where $\Rightarrow$ denotes convergence in distribution. Then $X_n \Rightarrow X$, but $\text{Var}(X_n) = \text{Var}(X) + \text{Var}(\xi_n) > \text{Var}(X)$.
The variance of $\xi_n$ can be made constant while preserving the property that $\xi_n \Rightarrow 0$. For instance, one can take $\xi_n = \pm\sqrt{n}$ with probability $1/2n$ and $0$ otherwise. Then $$\limsup_{n\rightarrow\infty} \text{Var}(X_n) = 1 + \text{Var}(X) > \text{Var}(X).$$