Given the IV estimator for $\beta_1$ (From $y = \beta_0 +\beta_1x + u$):
$$\beta_1 = \frac{Cov(z, y)}{Cov(z, x)}$$
And it's sample analog:
$$\hat{\beta_1} = \frac{\hat{Cov}(z, y)}{\hat{Cov}(z, x)}$$
Which simplifies to:
$$\hat{\beta_1} = \frac{\sum_{i=1}^n(z_i-\bar{z})(\beta_0 +\beta_1x_i + u)}{\sum_{i=1}^n(z_i - \bar{z})(x_i - \bar{x})} \rightarrow \beta_1 + \frac{\sum_{i=1}^n(z_i - \bar{z})u_i}{\sum_{i=1}^n(z_i - \bar{z})(x_i - \bar{x})} $$
Surely this estimator is unbiased because $Cov(z, u)=0$ by the assumption of exogeneity when dealing with instruments? Even if the above represents the sample covariance, because the sample covariance is an unbiased estimator of the population covariance, then surely $\hat{\beta_1}$ is still unbiased?
Aside from that, why is it necessary to take the conditional expectation on both $X$ and $Z$?.
$$E[\hat{\beta_1}] = E[E[\hat{\beta_1}|X, Z]] \rightarrow \frac{\sum_{i=1}^n(z_i - \bar{z})E[u_i|X, Z]}{\sum_{i=1}^n(z_i - \bar{z})(x_i - \bar{x})}$$
I can see why this expression isn't equal to zero (Because of the implication of the conditional on $X$), but why do we have to take that expectation instead of simply $E[\hat{\beta_1}] = E[E[\hat{\beta_1}|X]] $ or $E[E[\hat{\beta_1}|Z]]$? Is it necessary to condition on all of the data within the model?