How do you prove Cov $\left( \bar{X} , X_i - \bar{X} \right) = 0$?
Hint: Use properties of the covariance function and independence of $X_1,...,X_n$.
Solution:
Note that $$ Cov(\bar{X},X_i-\bar{X}) = Cov(\bar{X},X_i)-Cov(\bar{X},\bar{X}) = Cov(\bar{X},X_i)-Var(\bar{X}) $$
You know that $Var(\bar{X})=\frac{\sigma^2}{n}$. So you need to find $Cov(\bar{X},X_i)$. But,
$$ Cov(\bar{X},X_i)= Cov(\frac{1}{n}\sum_{j=1}^{n}X_j,X_i) = \frac{1}{n} \sum_{j=1}^{n} Cov(X_j,X_i) $$ Since $X_j$'s are independent it follows that $$Cov(X_j,X_i)=0 \text{ for all j } \neq i $$
So we conclude that $$ Cov(\bar{X},X_i) = \frac{1}{n} Cov(X_i,X_i) = \frac{\sigma^2}{n} $$
It follows that $$Cov(\bar{X},X_i-\bar{X}) = 0$$