[proofplan] We prove the theorem by strong induction on $n$. For the inductive step, let $B$ be the block containing $1$ and write the complement of $B$ as the linear gaps between consecutive elements of $B$, together with the final gap after the largest element of $B$. Noncrossing forces every block that meets one of these gaps to be contained entirely in that gap, so any nonempty gap inherits a smaller noncrossing partition. Applying the induction hypothesis inside such a gap gives an interval block in the gap, hence an interval block in the original partition; since the gap omits $1$, this block is proper in $[n]$. [/proofplan]

[step:Start the strong induction] For $n=1$, the only partition of $[1]$ is $1_1=\{\{1\}\}$. Its unique block is $\{1\}$, and $\{1\}=\{p,p+1,\dots,q\}$ with $p=q=1$. Hence $\{1\}$ is an interval block. Since this partition is $1_1$, the proper-block assertion is vacuous. [/step]

[step:Split the complement of the block containing $1$ into gaps] Assume $n\geq 2$, and assume the theorem has been proved for every positive integer smaller than $n$. Let $B$ denote the block of $\pi$ containing $1$. Write the elements of $B$ in increasing order as $B=\{b_1,\dots,b_r\}$, where $r=|B|$ and $1=b_1<b_2<\cdots<b_r\leq n$. If $B=[n]$, then $B=\{1,2,\dots,n\}$ is an interval block. In this case the partition has only the block $[n]$, so $\pi=1_n$, and the proper-block assertion is vacuous. We may therefore assume from now on that $B\neq [n]$. For each $s\in\{1,\dots,r-1\}$, define the internal gap $G_s$ by $G_s=\{b_s+1,b_s+2,\dots,b_{s+1}-1\}$. Define the final gap $G_r$ by $G_r=\{b_r+1,b_r+2,\dots,n\}$. Empty sets are allowed in this notation. Since $B\neq [n]$, at least one of the sets $G_1,\dots,G_r$ is nonempty. [/step]

[step:Show that any block meeting a gap is contained in that gap]Let $G_s$ be one of the gaps defined above, and let $C$ be a block of $\pi$ such that $C\cap G_s\neq\varnothing$. Since $G_s\cap B=\varnothing$, the blocks $B$ and $C$ are distinct. Choose an element $c\in C\cap G_s$. First suppose $s<r$. We prove $C\subseteq G_s$. If there were an element $d\in C$ with $d<b_s$, then $d<b_s<c<b_{s+1}$, where $d,c\in C$ and $b_s,b_{s+1}\in B$. This is a crossing pattern, contradicting that $\pi$ is noncrossing. If there were an element $d\in C$ with $d>b_{s+1}$, then $b_s<c<b_{s+1}<d$, where $b_s,b_{s+1}\in B$ and $c,d\in C$. This is again a crossing pattern. Because $G_s=\{x\in[n]: b_s<x<b_{s+1}\}$, no element of $C$ can lie outside $G_s$, so $C\subseteq G_s$. Now suppose $s=r$. We prove $C\subseteq G_r$. If there were an element $d\in C$ with $d<b_r$, then $d\neq 1$ because $1\in B$ and $B\cap C=\varnothing$. Hence $1<d<b_r<c$, where $1,b_r\in B$ and $d,c\in C$. This is a crossing pattern, contradicting that $\pi$ is noncrossing. Since $G_r=\{x\in[n]: b_r<x\leq n\}$, it follows that $C\subseteq G_r$.[/step]

[guided]We prove the containment carefully because this is the point where noncrossing is used. Let $B$ be the block of $\pi$ containing $1$, and write $B=\{b_1,\dots,b_r\}$ with $1=b_1<\cdots<b_r$. Fix a gap $G_s$, and let $C$ be a block with $C\cap G_s\neq\varnothing$. Choose $c\in C\cap G_s$. Since every gap is disjoint from $B$, the blocks $B$ and $C$ are distinct. Consider first an internal gap $G_s=\{b_s+1,\dots,b_{s+1}-1\}$ with $s<r$. The elements $b_s$ and $b_{s+1}$ are consecutive elements of the same block $B$, and the element $c$ lies strictly between them. Suppose that $C$ also contained some element $d$ outside this gap. If $d<b_s$, then the four elements occur in the order $d<b_s<c<b_{s+1}$. The first and third of these four elements, $d$ and $c$, lie in the same block $C$, while the second and fourth, $b_s$ and $b_{s+1}$, lie in the same block $B$. This is exactly the forbidden interlacing configuration for a noncrossing partition. If instead $d>b_{s+1}$, then the order is $b_s<c<b_{s+1}<d$. Now the first and third elements lie in $B$, and the second and fourth lie in $C$, again giving a forbidden crossing. Thus no element of $C$ can lie to the left of $b_s$ or to the right of $b_{s+1}$. Since $G_s$ consists precisely of the elements of $[n]$ strictly between $b_s$ and $b_{s+1}$, we obtain $C\subseteq G_s$. It remains to handle the final gap $G_r=\{b_r+1,\dots,n\}$. Again choose $c\in C\cap G_r$, so $b_r<c$. Suppose that $C$ contained an element $d<b_r$. Because $1\in B$ and $B\cap C=\varnothing$, we have $d\neq 1$. Therefore $1<d<b_r<c$. The elements $1$ and $b_r$ lie in the block $B$, while $d$ and $c$ lie in the block $C$. This is a forbidden crossing pattern. Hence $C$ has no element below $b_r$, and every element of $C$ lies in $G_r$.[/guided]

[step:Restrict the partition to a nonempty gap and relabel it] Choose a nonempty gap and denote it by $G$. By the previous step, every block of $\pi$ that meets $G$ is contained in $G$. Therefore the family $\rho:=\{C\in\pi:C\subseteq G\}$ is a partition of $G$. Let $m:=|G|$. Since $G$ is a nonempty proper subset of $[n]$, we have $1\leq m<n$. Let $a:=\min G$. Because $G$ is one of the gaps above, it has the form $G=\{a,a+1,\dots,a+m-1\}$. Define the increasing bijection $\varphi:[m]\to G$ by $\varphi(j)=a+j-1$ for $j\in[m]$. Pull back $\rho$ to a partition $\rho'$ of $[m]$ by declaring that $A\subseteq[m]$ is a block of $\rho'$ exactly when $\varphi(A)$ is a block of $\rho$. The partition $\rho'$ is noncrossing. Indeed, if $i_1<i_2<i_3<i_4$ in $[m]$ formed a crossing for two blocks of $\rho'$, then $\varphi(i_1)<\varphi(i_2)<\varphi(i_3)<\varphi(i_4)$ would form the same crossing for the corresponding two blocks of $\pi$, contradicting that $\pi$ is noncrossing. [/step]

[step:Apply the induction hypothesis inside the gap] Since $1\leq m<n$ and $\rho'$ is a noncrossing partition of $[m]$, the induction hypothesis gives an interval block $A$ of $\rho'$. Thus there exist integers $p$ and $q$ with $1\leq p\leq q\leq m$ such that $A=\{p,p+1,\dots,q\}$. Let $I:=\varphi(A)$. By the definition of the pullback partition, $I$ is a block of $\rho$, hence a block of $\pi$. Since $\varphi(j)=a+j-1$, we have $I=\{a+p-1,a+p,\dots,a+q-1\}$. Therefore $I$ is an interval block of $\pi$ in the original linear order on $[n]$. Finally, $I\subseteq G$ and $1\notin G$, so $I\neq[n]$. Hence $I$ is a proper subset of $[n]$. This proves the proper interval-block assertion whenever $\pi\neq 1_n$, and in particular proves the existence of an interval block. The strong induction is complete. [/step]