You probably mean the case $x_m \ne y_n$ and $z_k \ne y_n$ and $z_k \ne x_m$. Then $Z$ is LCS of $X_{m-1}$ and $Y_{n-1}$. For example, X=(2,1,3,6,4) and Y=(4,2,8,3,6,5), and Z=(2,3,6). Here, $5 \ne 6$ and $4 \ne 6$.So Z is LCS of (2,1,3,6) and (4,2,8,3,6). But this case can be considered as one of the previous cases: second or third.

$x_m \ne y_n$ and then $z_k \ne x_m \ne y_n$ is also repetition of assumption, $x_m\ \ne y_n$ two times?

You assume $x_m=y_n$ and then $z_k \ne x_m \ne y_n$. How is it possible $x_m=y_n$ and $x_m \ne y_n$?

@sprajagopal you start with n=1 and and simulate every possible string s in a systematic way (no string is missed). Assume such two strings exists s11 and s45 that is, M halts on s11 in, say, 2222 steps and M halts on s45 in 8882 steps. Then when we reach n=8882 our alg simulate s11 for 8882 steps and s45 for 8882. It is clear that M halts on both s11 and s45 in 8882 steps. So we accept <M>. But if such two s does not exists then our simulation never stops and hence never accepts <M>. That is definition of r.e. sets.