Let $P$ be a finite set with partial orderings $\leq_1$ and $\leq_2$. Then, there is a so-called dominating antichain $A \subseteq P$ where no two distinct elements of $A$ are comparabale in $\leq_1$ or $\leq_2$ and for every $p \in P$ there is an element $a \in A$ such that $p \leq_1 a$ or $p \leq_2 a$.
My approach:
Define $M \subseteq P$ as set of all $\leq_1$-maximal elements of $P$. Let $A \subseteq M$ be the set of all $\leq_2$-maximal elements of $M$.
Then, it is evident that no two distinct elements of $A$ are comparabale in $\leq_1$ or $\leq_2$.
Let $p \in P$.
Case 1: If $p \in A$, then $p \leq_1 p$.
Case 2: If $p \in M\setminus A$, then there is an element $a \in A$ with $p \leq_2 a$.
Case 3: If $p \in P\setminus M$, then - I do not know.
Do you have any advice for me? Thank you for your time and your insights!
Your construction of the dominating antichain $A$ is flawed. Let $P = \{a,b,c,d\}$, and let $b \leq_2 a$, $c \leq_1 a$, $d \leq_1 b$. Then, $M = \{a,b\}$ and $A = \{a\}$. But there is no element $x$ in $A$ such that $d \leq_1 x$ or $d \leq_2 x$.
Starting with the $\leq_1$-antichain $M$ is a good idea. Whenever you find two distinct elements $x,y$ in $M$ such that $y \leq_2 x$, you may modify $M$ as follows:
$$ M \gets M - \{y\} + \{z \in P \mid\ z \lessdot_1 y\ \land\ \lnot \exists m \in M: z \leq_1 m \}$$
Can you prove that this process terminates? This may lead to an inductive proof of the statement.