The down-set and up-set chain covers from a suitable maximum antichain splice to a width-sized chain cover

The antichain $A$ has cardinality $w$ and lies in both $P^{-}$ and $P^{+}$. Since every antichain of either induced subposet is also an antichain of $P$, both $P^{-}$ and $P^{+}$ have width exactly $w$.

Both induced subposets are nonempty because they contain $A$. They are proper subsets of $P$: indeed, $P^{-}=P$ would force $A$ to be exactly the set of maximal elements, since every maximal element of $P$ must then lie in $A$ and no $a\in A$ can have an element strictly above it. Dually, $P^{+}=P$ would force $A$ to be exactly the set of minimal elements. By finiteness, each therefore has cardinality strictly smaller than $\mathrm{\mid }P\mathrm{\mid }$.

By the hypothesis and step 1.1, choose chain covers $\{C_a^{-}:a\in A\}$ of $P^{-}$ and $\{C_a^{+}:a\in A\}$ of $P^{+}$, indexed so that $a\in C_a^{-}\cap C_a^{+}$. Such indexing is possible because each cover has $w=\mathrm{\mid }A\mathrm{\mid }$ chains, each chain contains at most one member of $A$, and all members of $A$ must be covered.

Fix $a\in A$. If $x\in C_a^{-}\subseteq P^{-}$ and $a<x$, then $x\le b$ for some $b\in A$, so $a<b$, contradicting that $A$ is an antichain. Hence every $x\in C_a^{-}$ satisfies $x\le a$. Dually, every $y\in C_a^{+}$ satisfies $a\le y$. Thus $x\le a\le y$ by transitivity, so $C_a^{-}\cup C_a^{+}$ is a chain.

The $w$ chains $C_a^{-}\cup C_a^{+}$ cover $P^{-}\cup P^{+}=P$. Thus they form the required width-sized chain cover of $P$.