Prove that if $p_1 \times p_2$ is positive, then $p_1$ is clockwise from $p_2$?
In two-dimensional space, $p_1 \times p_2$ is equal to $|p_1||p_2|\sin\theta$, where $\theta$ is the angle from $p_1$ to $p_2$. Depending which textbook you're using, this is probably the definition of the two-dimensional cross product—I'm not sure what definition Introduction to Algorithms uses.

$|p_1|$ and $|p_2|$ both have to be non-negative (by the definition of vector magnitude), so if neither of them is zero, the sign of the whole expression will be the sign of $\sin\theta$.

Now, how do we define "clockwise"? The easiest definition I know is is, $p_1$ is clockwise from $p_2$ if the angle(*) from $p_1$ to $p_2$ is less than 180 degrees. Try a few examples and you'll see how this definition works—remember that angles are traditionally measured with counter-clockwise being positive.

Now, we know that if $p_1 \times p_2$ is positive, then $\sin\theta$ must be positive. And if $\sin\theta$ is positive, then $\theta$ must be less than 180 degrees(*). So by the definition of clockwise, $p_1$ is clockwise from $p_2$.

(*) I'm assuming without loss of generality that all angles are positive and less than 360 degrees.