Verification wanted: Show the language $L=\{0^m1^n \enspace | \enspace m \neq n\}$ is not regular
This is a very hard one to prove via the pumping lemma, as you have to construct a string that eventually gives you something of the form $0^{n}1^{n}$ for some $n$ when you pump it the right number of times. As Renato notes, your decomposition (at the time of writing this answer) is incorrect. We'll come back to this.

A simpler way is to use the closure properties of regular languages. In particular, regular languages are closed under complementation and intersection, so


If $R$ is regular, then $\overline{R} = \Sigma^{\ast}\setminus R$ is also regular.
If $R_{1}$ and $R_{2}$ are regular, then $R_{3} = R_{1} \cap R_{2}$ is also regular.


So we assume that $L = \{0^{m}1^{n} \mid m \neq n\}$ is regular. Then by (1), $\overline{L} = \Sigma^{\ast}\setminus L$ is regular. In particular note that $\overline{L}$ includes all strings of the form $0^{n}1^{n}$. 

The language $A = \{0^{a}1^{b} \mid a,b \in \mathbb{N}\}$ is regular (we can show this via a regular expression, DFA or regular grammar for the language, e.g. $0^{\ast}1^{\ast}$).

Then by (2) $\overline{L}\cap A$ must also be regular, but this is the language $\{0^{n}1^{n}\}$, which is not regular (and much easier to prove via the pumping lemma!).

As noted in the question you linked however, it is possible to do it with the pumping lemma, but you need to pick your starting string carefully. As we only have two sections to the string, the obvious approach is to have a string of the form $0^{p}1^{x}$ for some $x$, so then we know that $y = 0^{k}$ for some $k\leq p$. Now the trick is to pick $x$ such that we can pump the string some number $y$ times and get $p + (y-1)\cdot k = x$. Of course we don't know what $k$ is - it could be anything from $1$ to $p$, so $x$ needs to take this into account. Hence $x$, in part, needs to (almost) be a multiple of every number from $1$ to $p$. So if we pick $x = p + p!$, we can say that for every $k$, there exists a $y$ such that $p + (y-1)\cdot k = p + p!$. In particular $y = \frac{p!}{k}+1$.

So then our starting string is $s=0^{p}1^{p+p!}$. By the pumping lemma if $L$ is regular, we can break $s$ up in $xyz$ such that $|y| \geq 1$, $|xy| \leq p$ and $xy^{i}z \in L$ for every $i \in \mathbb{N}$, but we have just shown that for $i = \frac{p!}{k}+1$, we get the string $0^{p+p!}1^{p+p!} \notin L$. Hence $L$ violates the pumping lemma, can thus can't be regular.