I've seen this proof in a textbook
$ E(D^*)=E(\frac{\sum_{k=1}^{n} (X_k - \bar{X})^2}{n})=E(\frac{\sum_{k=1}^{n} ((X_k - a) - (\bar{X} - a))^2}{n})$
then it simplifies to $\frac{E(\sum_{k=1}^{n} (X_k - a)^2)}{n}-E(\bar{X}-a)^2=\frac{n-1}{n}D\xi$
Please, explain the transition and why $E(\bar{X}-a)^2=\frac{D\xi}{n}$
Expand the square and note that $E\sum\limits_{k=1}^{n}(X_k-a)(\overset {-} {X}-a)=nE(\overset {-} {X}-a)^{2}$ because $\sum\limits_{k=1}^{n}(X_k-a)=n(\overset {-} {X}-a)$. Now you should be able to complete the argument for your first question. For the second question use the fact that the variance of a sum of independent random variables is the sum of the variances.