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?