Let $X_1, X_2, ..., X_n \sim U[0, 1]$ i.i.d. I want to derive $\textrm{Var}[F_n(x)]$ where $Y := F_n(x)$ is empirical cdf in point $x$.
My work so far
$$E[Y] = E[\frac 1 n \sum_{i = 1}^n 1_{\{X_i \le x\}}] = \frac 1 n \sum_{i = 1}^n F_X(x) = F_X(x)$$
Where we used fact that all the variables are independent and come from one distirbution $X$.
$$E[Y^2] = \frac{1}{n^2} \sum_{k = 1}^n \sum_{l = 1}^n E[1_{\{X_k \le x\}}1_{\{X_l \le x\}}] = \frac{1}{n^2}\sum_{k = 1}^n \sum_{l = 1}^n P(X_k \le x, X_l \le x)$$
Now term $P(X_k \le x, X_l \le x) = P(X_k \le x)P(X_l \le x)$ when $l \neq k$ due to independence, and $P(X_k \le x, X_l \le x) = P(X_k \le x)$ when $l = k$.
$$\frac{1}{n^2}\sum_{k = 1}^n \sum_{l = 1}^n P(X_k \le x, X_l \le x) = \frac{1}{n^2}\sum_{k = 1}^nP(X_k\le x)P(X_l \le x) + P(X_k \le x) = $$ =$$\frac{1}{n^2}\sum_{i = 1}^nP(X_k \le x)[(n - 1)F_X(x)] + P(X_k \le x) = $$ $$= \frac{1}{n^2}F_X(x) \cdot n \cdot [(n - 1)F_x + 1] = \frac 1 n F_X(x) \cdot [(n - 1)F_X(x) + 1]$$
now I want to ask you where is the mistake in my calculations, because I know that this variance should be equal to $F_X(x)(1 - F_X(x))$ which is different what I got.