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 $|P|$.

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 $w=|A|$ 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 $P$.