First let us remember the definitions for maximum and maximal elements.
Let $(P,<)$ be a partially ordered set, $x\in P$ is called maximal if for all $y\in P$ we have $y\ge x\rightarrow y=x$. In addition to that, $x$ is called a maximum if $x\ge y$ for all $y\in P$.
Note that being a maximum is stronger than being the unique maximal element, and one can easily concoct a partial order without a maximum, but with a unique maximal element.
Zorn's lemma: Let $(P,<)$ be a partially ordered set. If every chain is bounded, then there exists a maximal element. (It need not be a maximum.)
It seems that you try to show that if we have a maximum in $P$, and we remove it then we have a new maximal element (not necessarily a maximum though).
It would seem to me that you example you have in mind is a partially-well ordered set. However this is not always the case.
For example, $P=[0,1]$ with $<$ as the usual ordering of real numbers. We have that $1$ is the maximum element, however there is no "smallest" maximal element in $[0,1)$ under the same ordering.
This is because you have to require that every chain is bounded, if we have an unbounded chain whose only bound is the maximum then by removing it the partial order no longer fulfills the needed assumption of Zorn's lemma, and so it does no longer guarantee us a maximal element.