Prove that OLS estimator of the intercept has minimum variance 
Let $$y_i=B_0+B_1X_i+\varepsilon_i$$ where $\varepsilon_i\sim
 N(0,\sigma^2)$. Find the least squares estimator of $B_0$ and show that
   it is unbiased and has minimum variance.

I will not write in detail all the steps I went through, but
$$\hat{B_1}=\frac{\sum((X_i-\overline{X})(Y_i-\overline{Y})}{\sum(X_i-\overline{X})^2}$$
and
$$\hat{B_0}=\overline{Y}-\hat{B_1}\overline{X} \,.$$
Taking the expectation:
$$\mathbb{E}[\hat{B_0}]=\mathbb{E}[\overline{Y}-\hat{B_1}\overline{X}]=\mathbb{E}[\overline{Y}]-\overline{X}\mathbb{E}[\hat{B_1}]=B_0+B_1\overline{X}-B_1\overline{X}=B_0$$
then this is unbiased.
But how can I show that the estimator has minimum variance in this case?
EDIT: Since I already proved that $B_0$ is unbiased, and since the distribution of $B_0$ belongs to an exponential family, it's a complete and sufficient statistic. Thus this estimator has minimum variance by the Lehmann–Scheffé theorem.
 A: For establishing a more general result, I am referring to the lecture notes here.
Suppose we have the multiple linear regression model 
$$y=X\beta + \varepsilon$$
, where the design matrix $X$ (with non-random entries) of order $n\times k$ has full column rank and $\beta=(\beta_1,\ldots,\beta_k)^T$ is the vector of regression coefficients. 
Further assume that $\varepsilon \sim N_n(0,\sigma^2 I_n)$ with $\sigma^2$ unknown, so that $y\sim N_n(X\beta,\sigma^2 I_n)$.
Under this setting, we know that the OLS estimator of $\beta$ is $$\hat\beta=(X^T X)^{-1}X^T y\sim N_k\left(\beta,\sigma^2(X^T X)^{-1}\right)$$
So, $$(y-X\hat\beta)^T X(\hat\beta-\beta)=(y^TX-y^T X)(\hat\beta-\beta)=0$$
Hence,
\begin{align}
\left\| y-X\beta \right\|^2 &=\left\|y-X\hat\beta+X\hat\beta-X\beta\right\|^2
\\&=\left\|y-X\hat\beta\right\|^2+\left\|X\hat\beta-X\beta\right\|^2
\\&=\left\|y-X\hat\beta\right\|^2+\left\|X\hat\beta\right\|^2+\left\|X\beta\right\|^2-2\beta^T X^T y 
\end{align}
The pdf of $y$ now looks like
\begin{align}
f(y;\beta,\sigma^2)&=\frac{1}{(2\pi\sigma^2)^{n/2}}\exp\left[-\frac{1}{2\sigma^2}\left\| y-X\beta \right\|^2\right]
\\\\&=\frac{1}{(2\pi\sigma^2)^{n/2}}\exp\left[\frac{1}{\sigma^2}\beta^T X^T y-\frac{1}{2\sigma^2}\left(\left\|y-X\hat\beta\right\|^2+\left\|X\hat\beta\right\|^2\right)-\frac{1}{2\sigma^2}\left\|X\beta\right\|^2\right]
\end{align}
Noting that $X\hat\beta$ is a function of $X^T y$ and that the above density is a member of the exponential family, a complete sufficient statistic for $(\beta,\sigma^2)$ is given by $$T=\left(X^Ty,\left\|y-X\hat\beta\right\|^2\right)$$
Again $\hat\beta$, a function of $X^T y$, is also a function of $T$ and it is unbiased for $\beta$. 
So by Lehmann-Scheffe theorem $\hat\beta$ is the UMVUE of $\beta$.
