We proceed by induction on
\(n\text{.}\) In the base case
\(n=1\text{,}\) \(2^{[n]}\) itself is a symmetric chain.
Now, suppose that \(n\geq2\) and, using the inductive hypothesis, let \(\mathcal{C}_1,\mathcal{C}_2,\dots,\mathcal{C}_N\) be a collection of symmetric chains in \(2^{[n-1]}\) which partition it. For each \(m\) such that \(1\leq m\leq N\text{,}\) let
\begin{equation*}
k_m:=\min\{|A|: A\in \mathcal{C}_m\}
\end{equation*}
and write \(\mathcal{C}_m=\{X_{m}^i:
k_m\leq i\leq n-1-k_m\}\) where
\begin{equation*}
X_{m}^{k_m}\subseteq X_{m}^{k_m+1}\subseteq\cdots\subseteq X_{m}^{n-1-k_m}
\end{equation*}
and
\begin{equation*}
|X_{m}^i|=i\text{.}
\end{equation*}
For each \(1\leq m\leq N\text{,}\) define
\begin{equation*}
\mathcal{C}_m':=\{X_{m}^{k_m},X_{m}^{k_m+1},\dots,X_{m}^{n-1-k_m},X_{m}^{n-1-k_m}\cup\{n\}\}\text{ and }
\end{equation*}
\begin{equation*}
\mathcal{C}_m'':=\{X_{m}^{k_m}\cup\{n\},X_{m}^{k_m+1}\cup\{n\},\dots,X_{m}^{n-2-k_m}\cup\{n\}\}\text{.}
\end{equation*}
Each of the collections
\(\mathcal{C}_m'\) and
\(\mathcal{C}_m''\) is a symmetric chain in
\(2^{[n]}\) and every element of
\(2^{[n]}\) is contained in exactly one of
\(\mathcal{C}_1',\dots,\mathcal{C}_N',\mathcal{C}_1'',\dots,\mathcal{C}_N''\) (see
Exercise 2). This completes the proof.
