Prove that $\frac{1}{n(n-1)}\sum_{i=1}^{n}(X_{i} - \overline{X})^{2}$ is an unbiased estimate of $\text{Var}(\overline{X})$ If $X_{1},X_{2},\ldots,X_{n}$ are independent random variables with common mean $\mu$ and variances $\sigma^{2}_{1},\sigma^{2}_{2},\ldots,\sigma^{2}_{n}$. Prove that
\begin{align*}
\frac{1}{n(n-1)}\sum_{i=1}^{n}(X_{i} - \overline{X})^{2}
\end{align*}
is an unbiased estimate of $\text{Var}[\overline{X}]$.
MY ATTEMPT
To begin with, I tried to approach the problem as it follows
\begin{align*}
\text{Var}(\overline{X}) = \text{Var}\left(\frac{X_{1} + X_{2} + \ldots + X_{n}}{n}\right) = \frac{\sigma^{2}_{1} + \sigma^{2}_{2} + \ldots + \sigma^{2}_{n}}{n^{2}}
\end{align*}
But I do not know how to proceed from here. Any help is appreciated.

Let's minimize the algebra.  We can do this by focusing on the coefficient of $\sigma_1^2.$
First, because all the variables have the same mean,
$$E[X_i - \bar X]=\mu - \mu = 0$$
for all $i,$ implying
$$\operatorname{Var}(X_i - \bar X) = E[(X_i - \bar X)^2] - E[X_i - \bar X]^2 = E[(X_i - \bar X)^2].$$
Since the sum of the $(X_i-\bar X)^2$ is invariant under any permutation of the indexes, let's study the case $i=1$ because it will show us what happens with all the $i.$
We can  easily split $X_1 - \bar X$ into independent (and therefore uncorrelated) parts as
$$X_1 - \bar X = X_1 - \frac{1}{n}\left(X_1 + X_2 + \cdots + X_n\right) = \frac{n-1}{n}\left(X_1 - \frac{1}{n-1}\left(X_2 + \cdots + X_n\right)\right).$$
Taking variances immediately gives
$$\operatorname{Var}(X_1 - \bar X) = \left(\frac{n-1}{n}\right)^2 \left(\sigma_1^2 + \frac{1}{(n-1)^2}\left(\sigma_2^2 + \cdots + \sigma_n^2\right)\right).$$
(This is the payoff from observing that the expectation of the square of $X_i - \bar X$ is its variance.)
When summing over all $i$ and ignoring the common factor of $((n-1)/n)^2,$ $\sigma_1^2$ will therefore appear once and it will appear $n-1$ more times with a factor of $1/(n-1)^2,$ for a total coefficient of
$$1 + (n-1)\left(\frac{1}{(n-1)^2}\right) = \frac{n}{n-1}.$$
Consequently every $\sigma_i^2$ appears with this coefficient, whence
$$\eqalign{
E \Bigg[ \frac{1}{n(n-1)}\sum_{i=1}^{n}(X_i - \bar{X})^{2} \Bigg]
&= \frac{1}{n(n-1)} \sum_{i=1}^{n}\operatorname{Var}(X_i - \bar{X}) \\
&= \frac{1}{n(n-1)} \left(\frac{n-1}{n} \right)^2 \left[  \frac{n}{n-1} \sum_{i=1}^n \sigma_i^2 \right] \\
&= \frac{\sigma_1^2 + \sigma_2^2 + \cdots + \sigma_n^2}{n^2},
}$$
QED.
