The Erdős-Szekeres monotone subsequence theorem follows by applying Mirsky's theorem to the index-value poset

If $r=1$ or $s=1$, any one-term sublist has the required kind, so assume $r,s\ge2$.

On the index set $\{0,\ldots,N-1\}$ define $i\preceq j$ when $i\le j$ and $a_i\le a_j$. The relation is reflexive and transitive componentwise, while $i\preceq j\preceq i$ forces $i=j$, so it is a partial order. Its chains, read in increasing index order, give strictly increasing sublists because the values are pairwise distinct.

Suppose there is no strictly increasing sublist of length $r$. Then the index-value poset has height at most $r-1$, so [L1] covers its indices by at most $r-1$ antichains. After ordering the covering antichains and removing from each one the indices already assigned to an earlier one, they form a partition into at most $r-1$ antichains.

In an antichain of the index-value poset, increasing the indices strictly decreases the corresponding values: if $i<j$ then $a_i<a_j$ would make $i\prec j$, while equality is excluded. Hence, if there is no strictly decreasing sublist of length $s$, every such antichain has at most $s-1$ members.

Under the simultaneous absence of both required sublists, map each index to the part containing it in the partition from step 2.1. There are at most $r-1$ parts, and step 2.2 says that every fibre has at most $s-1$ elements. Thus [L2] gives $N\le(r-1)(s-1)$, contradicting $N=(r-1)(s-1)+1$.

Therefore at least one of the two sublists exists: a strictly increasing one of length $r$, or a strictly decreasing one of length $s$.