A convex chain (in particular a covering pair) of a finite poset occurs consecutively in some linear extension

Setup. Enumerate the nonempty chain $C$ in increasing order as $C=\{c_1\prec c_2\prec \cdots \prec c_m\}$, and set $Q:=(P\setminus C)\cup \{\ast \}$ for a new element $\ast \mathrm{\notin }P$. Let $R$ be the relation on $Q$ consisting of the pairs $(x,y)$ with $x,y\in P\setminus C$ and $x\prec y$, the pairs $(x,\ast )$ with $x\in P\setminus C$ and $x\prec c$ for some $c\in C$, and the pairs $(\ast ,y)$ with $y\in P\setminus C$ and $c\prec y$ for some $c\in C$. Let ${⪯}_Q$ be the reflexive transitive closure of $R$, so ${⪯}_Q$ is reflexive and transitive by construction.

The relation ${⪯}_Q$ is antisymmetric. A cycle of $R$ whose vertices lie in $P\setminus C$ would produce $x_1\prec x_2\prec \cdots \prec x_1$ in the poset $P$, impossible by transitivity and antisymmetry; so every nontrivial cycle passes through $\ast$, and between two consecutive occurrences of $\ast$ it consists of an edge $(\ast ,y)$, a path inside $P\setminus C$ from $y$ to an element $x$, and an edge $(x,\ast )$. By the definition of $R$ there are then $c^{\mathrm{\prime }},c\in C$ with $c^{\mathrm{\prime }}\prec y$ and $x\prec c$, and the path inside $P\setminus C$ gives $y⪯x$; hence $c^{\mathrm{\prime }}⪯y⪯x⪯c$. Since $c^{\mathrm{\prime }},c\in C$ and $C$ is convex, this forces $x\in C$, contradicting $x\in P\setminus C$. Therefore $R$ has no nontrivial cycles, and ${⪯}_Q$ is a partial order on the finite set $Q$.

By [F4] the finite poset $Q$ has a linear extension ${\pi }_Q=(u_1,\dots ,u_r)$; let $\pi$ be the tuple obtained from ${\pi }_Q$ by replacing the one occurrence of $\ast$ with the block $(c_1,\dots ,c_m)$. Then $\pi$ lists every element of $P$ exactly once. It is a linear extension of $P$: if $x\prec y$ with $x,y\in P\setminus C$ then $x{⪯}_{Q}y$, so $x$ precedes $y$ in $\pi$; if $x\in P\setminus C$ and $x\prec c_i$ then $(x,\ast )\in R$, so $x$ precedes $\ast$ in ${\pi }_Q$ and hence precedes the whole block; if $c_i\prec y$ with $y\in P\setminus C$ then $(\ast ,y)\in R$, so the whole block precedes $y$; and the block itself lists $c_1,\dots ,c_m$ in increasing order, so it respects the relations inside $C$. Since $C$ exhausts the block, its elements occur consecutively in $\pi$.

In particular, let $x⋖y$ be a covering pair and put $C:=\{x,y\}$. Then $C$ is a nonempty chain, and it is convex: if $x⪯z⪯y$ with $z\in P$, then either $z=x$, or $z=y$, or $x\prec z\prec y$, which is excluded by [F3]; in all cases $z\in C$. So step 3.1 applies and yields a linear extension of $P$ in which $x$ and $y$ are consecutive.