Proof OLS is Biased if a Regressor is Correlated with the error Consider the basic linear model:
$$y_i =\beta_0 +\beta_1 x_i +\varepsilon_i$$
I am aware that if $E[\varepsilon_i|x_i ]=0$ then $E[\hat{\beta_1}]=\beta_1$ (unbiasedness) and also that if $Cov(x_i,\varepsilon_i)=0$ then $plim \hat{\beta_1} = \beta_1$ (consistency).
I am hoping to prove that if $Cov(x_i,\varepsilon_i) \ne 0$ then $E[\hat{\beta_1}]\ne\beta_1$.
Can anyone help? My attempt begins below.
$$\hat{\beta_1}=\frac{\sum_{i=1}^n (x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n (x_i-\bar{x})^2}$$
$$\hat{\beta_1}=\beta_1 + \frac{\sum_{i=1}^n (x_i-\bar{x})(\varepsilon_i-\bar{\varepsilon})}{\sum_{i=1}^n (x_i-\bar{x})^2}$$
$$E[\hat{\beta_1}]=\beta_1 + E\left[\frac{\sum_{i=1}^n (x_i-\bar{x})(\varepsilon_i-\bar{\varepsilon})}{\sum_{i=1}^n (x_i-\bar{x})^2}\right]$$
My goal is to show the expectation on the right must be non-zero as a result of $Cov(x,\varepsilon)\ne 0$. The difficulty is that the expectation operator can be separately applied to the numerator and denominator. The numerator is in expectation non-zero, but if the sample covariance of $x$ and $\varepsilon$ is not independent of the sample variance of $x$, I don't know how to proceed.
Thanks in advance!

You are almost there. Remember that $\bar{\epsilon}=0$, you get:
$$
\hat{\beta}_1=\beta+\frac{\sum_{i=1}^n(x_i-\bar{x})(\epsilon_i-\bar{\epsilon})}{\sum_{i=1}^n(x_i-\bar{x})^2}=\beta+\frac{\sum_{i=1}^n(x_i-\bar{x})\epsilon_i}{\sum_{i=1}^n(x_i-\bar{x})^2}
$$
Therefore
$$
E(\hat{\beta}_1 \vert x)=E\left(\left. \beta+\frac{\sum_{i=1}^n(x_i-\bar{x})\epsilon_i}{\sum_{i=1}^n(x_i-\bar{x})^2}\right|x\right)=\beta+\frac{\sum_{i=1}^n(x_i-\bar{x})E(\epsilon_i\vert x)}{\sum_{i=1}^n(x_i-\bar{x})^2}\neq \beta
$$
since $E(\epsilon_i \vert x)\neq 0$. Since, $E(\hat{\beta}_1 \vert x) \neq \beta$ it follows that $E(\hat{\beta}_1)=E(E(\hat{\beta}_1 \vert x))\neq \beta$.
