The fact that the divisor "$X\,$" is not ample can also be seen directly from the definition. For each point $p$ of $Y$ other than the image of the divisor "$X\,$", a section of $mX$ has no poles on $X \times p$, and is thus constant on $X \times p$. Therefore there is no $m$ for which the linear series $\Gamma(mX)$ separates the points of $X \times Y$. (In fact the map from $X \times Y$ to projective space associated with $\Gamma(mX)$ always factors through the projection $X \times Y \rightarrow Y$.

P.S. How did this question from 2012 make it to the top page?

The fact that the divisor "$X\,$" is not ample can also be seen directly from the definition, and does not depend on the characteristic.

As usual we may assume $k$ is algebraically closed, so that $Y$ has infinitely many points $p$.
Choose some $p$ other than the image of the divisor "$X\,$". Then $X \times p$ is a complete curve, and the restriction to $X \times p$ of any section of $mX$ has no poles and is therefore constant. Therefore there is no $m$ for which the linear series $\Gamma(mX)$ separates the points of $X \times Y$. (In fact the map from $X \times Y$ to projective space associated with $\Gamma(mX)$ always factors through the projection $X \times Y \rightarrow Y$.)

P.S. How did this question from 2012 make it to the top page?