In Fumio Hayashi's Econometrics, Lemma 2.1 (convergence in distribution and in moments) states that
Lemma 2.1 $\quad$ Let $\alpha_{sn}$ be the $s$-th moment of $z_n$ and $\lim_{n\to\infty}\alpha_{sn}=\alpha_s$ where $\alpha_s$ is finite. Then $$ z_n \to_d z \implies \alpha_s \text{ is the $s$-th moment of $z$}$$
Then the author goes on to comment that
For example, if the variance of a sequence of random variables converging in distribution converges to some finite number, then that number is the variance of the limiting distribution.
My question is, Why is this comment true? I don't think this lemma about moments can be directly applied to the variance, which is a central moment. If this is not a direct application, it remains unclear to me how this lemma can lead to the author's claim about the variance.
Call the sequence of our concern $y_n\to_d y$. To apply the lemma to the variance, we need to convert $y_n$ into $z_n:=y_n-\mathbb{E}[y_n]$, so that the convergence $\operatorname{Var}[y_n]\to\alpha$ for some $\alpha\in\mathbb{R}$ is translated into $\mathbb{E}[(z_n)^2] \to \alpha$. The obstacle to apply the lemma is, we do not know $z_n:=y_n-\mathbb{E}[y_n]$ converges in distribution to $y-\mathbb{E}[y]$ or not (for one thing, $\{\mathbb{E}[y_n]\}$ is not known to converge). If this were known, then by the lemma $\mathbb{E}[(y-\mathbb{E}[y])^2]=\alpha$, as desired.
As pointed out in the comments to the question, both Lemma 2.1 and Hayashi's comment are incorrect. The issue with the Lemma is in Hayashi's errata.
Counterexample to Lemma 2.1: Let $z_n$ be such that $\mathbb{P}\{z_n=n\} = \frac{1}{n}$ and $\mathbb{P}\{z_n=0\} = \frac{n-1}{n}$. Then $\mathbb{E}[z_n] = 1$. $z_n \to_d 0 =: z$. Clearly, $1=\lim_{n\to\infty}\mathbb{E}[z_n] \ne \mathbb{E}[z]=0.$
Counterexample to Hayashi's comment: Let $z_n$ be such that $\mathbb{P}\{z_n = \sqrt{n}\} = \mathbb{P}\{z_n = -\sqrt{n}\} = \frac{1}{2n}$ and $\mathbb{P}\{z_n = 0\} = 1-\frac{1}{n}.$ Then $\operatorname{Var}(z_n) = 1 \to 1$ as $n\to\infty.$ However, $z_n\to_d 0 =: z,$ so $\operatorname{Var}(z) = 0 \ne 1.$