As is shown in the underlined part of the picture above. First, why after $c^+ \not\in S^-$, we can use induction hypothesis? how does it imply $S^-$ contains no antichain of cardinal $m + 1$? Second, why if $x \in S_i^-$, then $x \leq a_j$ for some $j$? I believe $a_j \in A$, it seems $x \parallel a_j$ is possible as well.
Since $c^+\notin S^-$, we have $c^+\in S\setminus S^-$. The set $S$ is finite, so this implies that $|S^-|<|S|$, and the induction hypothesis therefore applies to $S^-$. (Remember, the induction is on $|S|$, not on $m$.) Every antichain in $S^-$ is an antichain in $S$, and $S$ has no antichain of cardinality $m+1$, so $S^-$ also has no antichain of cardinality $m+1$. Thus, the induction hypothesis does ensure that $S^-$ is the union of $m$ chains, say $S^-=S_1^-\cup\ldots\cup S_m^-$. Moreover, since $\{a_1,\ldots,a_m\}$ is an antichain in $S^-$, each $a_k$ must be in a different chain, and we might as well index the chains so that $a_k\in S_k^-$ for $k=1,\ldots,m$.
Now suppose that $x\in S_i^-$. Then $x\in S^-=\{y\in S:y\le a_j\text{ for some }j\}$, so $x\le a_j$ for some $j$; it’s just the definition of $S^-$.
You didn’t ask, but I’ll finish the argument, just to play safe. We know that $a_i\in S_i^-$, and $S_i^-$ is a chain, so either $a_i<x$, or $x\le a_i$. If $a_i<x$, then $a_i<a_j$, which is impossible, so $x\le a_i$; and $x$ was an arbitrary element of $S_i^-$, so $a_i$ must be the maximum element of $S_i^-$.