proof for Erdős-Szekeres theorem using Dilworth's theorem Let's review a few definitions:
Dilwoths's theorem: Suppose that the length of the longest antichain in the poset $P$ is $r$, then $P$ can be partitioned into $r$ chains.
Dilworth's dual theorem: Suppose that the length of the longest chain in the poset $P$ is $r$, then $P$ can be partitioned into $r$ antichains.
Erdős-Szekeres theorem: Given $n \geq rs+1$, any sequence of $n$ elements has either an increasing subsequence of length $r+1$ or a decreasing subsequence of lengthe $s+1$.
Now here's the proof:
Define a partial ordering on the members of the sequence, in which $x$ is less than or equal to $y$ in the partial order if $x ≤ y$ as numbers and $x $ is not later than $y$ in the sequence. A chain in this partial order is a monotonically increasing subsequence, and an antichain is a monotonically decreasing subsequence. By Dilworth's dual theorem, either there is a chain of length $r$, or the sequence can be partitioned into at most $r − 1$ antichains; but in that case the largest of the antichains must form a decreasing subsequence with length at least
   $$\lceil\frac{rs-r-s+2}{r-1} \rceil =s.$$
Alternatively, by Dilworth's theorem itself, either there is an antichain of length $s$, or the sequence can be partitioned into at most $s − 1$ chains, the longest of which must have length at least $r$.
What I really don't comprehend about this proof, is the relation between the definitions and the bolded part, which is supposedly deduced from the definitons.
Any help would be appreciated.
 A: The proof actually starts with a sequence of length $n\ge(r-1)(s-1)+1$ and shows that it has either an increasing subsequence of length $r$ or a decreasing subsequence of length $s$. (In other words, it shifts the $r$ and $s$ of the usual statement of the theorem by $1$.)
Suppose that there is no chain of length $r$ in the given partial order. Then the maximal length of any chain is at most $r-1$, and by the dual Dilworth theorem the partial order can be partitioned into at most $r-1$ antichains. The partial order has at least 
$$(r-1)(s-1)+1=rs-r-s+2$$
elements; if you divide these into $r-1$ or fewer parts, the mean part size is at least
$$\frac{rs-r-s+2}{r-1}=\frac{(r-1)(s-1)+1}{r-1}=s-1+\frac1{r-1}\;,$$
so the largest part must be at least
$$\left\lceil\frac{rs-r-s+2}{r-1}\right\rceil=\left\lceil s-1+\frac1{r-1}\right\rceil=s-1+\left\lceil\frac1{r-1}\right\rceil=s-1+1=s\;.$$
The second part of the argument is virtually identical, with the rôles of $r$ and $s$ interchanged and using Dilworth’s theorem rather than its dual.
A: Erdős-Szekeres theorem: Given $n≥rs+1$, any sequence of $n$ elements has either an increasing subsequence of length $r+1$ or a decreasing subsequence of length $s+1$.
As mentioned by Brian M. Scott, the bold part in your proof is a proof for an equivalent formulation:
given $n \geq (r-1)(s-1)+1$ for $r,s \in \mathbb{Z}_{>0}$, any sequence of $n$ distinct reals has either an increasing subsequence of length $r$ or a decreasing subsequence of length $s$.
To prove the original formulation ($n \geq rs + 1$): first define the same partially ordered set $P$ as you mentioned.
Case 1: there is a chain of length at least $r+1$ as desired.
Case 2: the longest chain has length $\ell \leq r$. By Dilworth's Dual Theorem (Mirsky's Theorem), we can partition $P$ into $k$ antichains, where $k=\ell \leq r$.
Suppose for contradiction that no antichain $A_i$ in the partition has length $|A_i| \geq s+1$. So $|A_i| \leq s$ for each $A_i$. So there can be at most $rs$ total elements in our antichain partition, since $\sum_{i=1}^k|A_i|\leq\sum_{i=1}^ks=ks\leq rs$. But our antichain partition must have $|P| = n \geq rs+1$ elements.
So some antichain in the partition has length at least $s+1$ as desired.
