We have $X_1,X_2,...$ following iid.
Let $\overline{X}_n=\frac{1}{n}\displaystyle \sum_{i=1}^n X_i$ and suppose that $E[X^4_1]$ has a finite value.
What are the values for $\displaystyle \lim_{ n \to \infty } E[(\overline{X}_n)^2] $ and $\displaystyle \lim_{ n \to \infty } V[(\overline{X}_n)^2] $ ?
Note that $$\begin{align}\mathbb{E}\{(\overline{X}_n)^2\}&=\mathbb{E}\left\{\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^2\right\}\\ &=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n\mathbb{E}\{X_iX_j\}\\ &=\frac{1}{n^2}\sum_{i=1}^n\mathbb{E}\{X_i^2\}+\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1\\j\neq i}^n\mathbb{E}\{X_i\}\mathbb{E}\{X_j\}\\ &=\frac{1}{n^2}\sum_{i=1}^n\mathbb{E}\{X_1^2\}+\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1\\j\neq i}^n\mathbb{E}\{X_1\}\mathbb{E}\{X_1\}\\ &=\frac{1}{n}\mathbb{E}\{X_1^2\}+\frac{n(n-1)}{n^2}\mathbb{E}\{X_1\}^2\\ \end{align}$$ where the second line comes from multiplying two sums, the third line comes from separating the cases when the sum indices $(i,j)$ are equal -yielding the second moment- or different -yielding the product of independent variables-, and finally replacing by $X_1$ because the $X_i$ are iid. You can take the limit here, as the second and first moment are finite given that the fourth moment is finite.
You can decompose the variance with a similar procedure.