The key thing here is that Wooldridge makes the assumption of random sampling.

Notice that since we have a random sample this means $(x_i, y_i) \perp (x_j, y_j)$ for $i \neq j$ which means that the single components of the pair are also independent, in particular $x_i \perp x_j$ (to see that just notice the joint is $p(x_i, y_i, x_j, y_j) = p(x_i, y_i)p (x_j, y_j)$ and marginalize over $y$).

This further implies $y_i \perp y_j |x_i, x_j$, since:

$$ p(y_i, y_j|x_i, x_j) = \frac{p(x_i, y_i, x_j, y_j)}{p(x_i, x_j)} = \frac{p(x_i, y_i)p (x_j, y_j)}{p(x_i)p(x_j)} = p(y_i|x_i)p(y_j|x_j) $$

But $y_i|x_i$ is nothing more than then the disturbance $u_i$ plus a constant. Hence $u_i \perp u_j |x$ when assuming a random sample and $E[u_i u_j | x] =E[u_i|x] E[u_j | x] = 0$

The key thing here is that Wooldridge makes the assumption of random sampling.

Notice that since we have a random sample this means $(x_i, y_i) \perp (x_j, y_j)$ for $i \neq j$ which means that the single components of the pair are also independent, in particular $x_i \perp x_j$ (to see that just notice the joint is $p(x_i, y_i, x_j, y_j) = p(x_i, y_i)p (x_j, y_j)$ and marginalize over $y$).

This further implies $y_i \perp y_j |x_i, x_j$, since:

$$ p(y_i, y_j|x_i, x_j) = \frac{p(x_i, y_i, x_j, y_j)}{p(x_i, x_j)} = \frac{p(x_i, y_i)p (x_j, y_j)}{p(x_i)p(x_j)} = p(y_i|x_i)p(y_j|x_j) $$

But $y_i|x_i$ is nothing more than the disturbance $u_i$ plus a constant. Hence $u_i \perp u_j |x$ when assuming a random sample and $E[u_i u_j | x] =E[u_i|x] E[u_j | x] = 0$

The key thing here is that Wooldridge makes the assumption of random sampling.

Notice that since we have a random sample this means $(x_i, y_i) \perp (x_j, y_j)$ which means that the single components of the pair are also independent, in particular $x_i \perp x_j$ (to see that just notice the joint is $p(x_i, y_i, x_j, y_j) = p(x_i, y_i)p (x_j, y_j)$ and marginalize over $y$).

This further implies $y_i \perp y_j |x_i, x_j$, since:

$$ p(y_i, y_j|x_i, x_j) = \frac{p(x_i, y_i, x_j, y_j)}{p(x_i, x_j)} = \frac{p(x_i, y_i)p (x_j, y_j)}{p(x_i)p(x_j)} = p(y_i|x_i)p(y_j|x_j) $$

But $y_i|x_i$ is nothing more than then the disturbance $u_i$ plus a constant. Hence $u_i \perp u_j |x$ when assuming a random sample and $E[u_i u_j | x] =E[u_i|x] E[u_j | x] = 0$

The key thing here is that Wooldridge makes the assumption of random sampling.

Notice that since we have a random sample this means $(x_i, y_i) \perp (x_j, y_j)$ for $i \neq j$ which means that the single components of the pair are also independent, in particular $x_i \perp x_j$ (to see that just notice the joint is $p(x_i, y_i, x_j, y_j) = p(x_i, y_i)p (x_j, y_j)$ and marginalize over $y$).

This further implies $y_i \perp y_j |x_i, x_j$, since:

$$ p(y_i, y_j|x_i, x_j) = \frac{p(x_i, y_i, x_j, y_j)}{p(x_i, x_j)} = \frac{p(x_i, y_i)p (x_j, y_j)}{p(x_i)p(x_j)} = p(y_i|x_i)p(y_j|x_j) $$

But $y_i|x_i$ is nothing more than then the disturbance $u_i$ plus a constant. Hence $u_i \perp u_j |x$ when assuming a random sample and $E[u_i u_j | x] =E[u_i|x] E[u_j | x] = 0$