[proofplan]
We prove that the interval indexed by $F_k$ consists exactly of the faces whose first containing facet in the shelling order is $F_k$. For a fixed facet $F_k$, we compare the faces inside $F_k$ with the subcomplex generated by the earlier facets. The shelling condition says that the old part of $F_k$ is generated by certain codimension-one faces, and this makes the new faces precisely those containing the restriction face $R(F_k)$. The first-appearance description then gives both disjointness and that every face is included.
[/proofplan]
[step:Describe the old faces inside a fixed facet using the shelling condition]
For $k\in\{1,\dots,t\}$, define the subcomplex generated by the first $k$ facets by
\begin{align*}
\Delta_k=\{G\subseteq F_i : 1\le i\le k\}.
\end{align*}
For $k=0$, set $\Delta_0=\varnothing$ as a bookkeeping convention for the empty previous stage. For every subset $S$ of the vertex set of $\Delta$, write $2^S=\{G:G\subseteq S\}$ for the full simplex on $S$.
Fix $k\ge 2$. A face $G\subseteq F_k$ is called old at stage $k$ if $G\in\Delta_{k-1}$. Define
\begin{align*}
A_k=\{v\in F_k : F_k\setminus\{v\}\in \Delta_{k-1}\}.
\end{align*}
Because $F_1,\dots,F_t$ is a shelling order, the intersection $\Delta_{k-1}\cap 2^{F_k}$ is the subcomplex of $2^{F_k}$ generated by the codimension-one faces $F_k\setminus\{v\}$ that already lie in $\Delta_{k-1}$. Therefore, for every $G\subseteq F_k$,
\begin{align*}
G\in\Delta_{k-1}
\end{align*}
if and only if
\begin{align*}
G\subseteq F_k\setminus\{v\}\text{ for some }v\in A_k.
\end{align*}
[guided]
We isolate exactly which faces of the new facet $F_k$ have already appeared before stage $k$. The previous stage is the subcomplex
\begin{align*}
\Delta_{k-1}=\{G\subseteq F_i : 1\le i<k\}.
\end{align*}
Thus a face $G\subseteq F_k$ is old precisely when $G\in\Delta_{k-1}$. For any subset $S$ of the vertex set of $\Delta$, the notation $2^S$ denotes the full simplex $\{G:G\subseteq S\}$.
The shelling condition is the key input. Applied to the facet $F_k$, it says that the intersection of the old complex with the simplex on $F_k$ is generated by codimension-one faces of $F_k$. We record exactly which codimension-one faces these are by defining
\begin{align*}
A_k=\{v\in F_k : F_k\setminus\{v\}\in \Delta_{k-1}\}.
\end{align*}
Here $F_k\setminus\{v\}$ is the codimension-one face obtained by deleting the vertex $v$ from the facet $F_k$.
Since $\Delta_{k-1}\cap 2^{F_k}$ is generated by the faces $F_k\setminus\{v\}$ with $v\in A_k$, a face $G\subseteq F_k$ belongs to $\Delta_{k-1}$ exactly when it is contained in one of those generating codimension-one faces. Hence, for every $G\subseteq F_k$,
\begin{align*}
G\in\Delta_{k-1}
\end{align*}
if and only if
\begin{align*}
G\subseteq F_k\setminus\{v\}\text{ for some }v\in A_k.
\end{align*}
This is the point where shellability is used: it converts the old part of $F_k$ into a union of codimension-one faces, so that newness can be tested vertex-by-vertex.
[/guided]
[/step]
[step:Identify the new faces of each facet with its restriction interval]
Fix $k\ge 2$ and let $G\subseteq F_k$. By the previous step, $G$ is new at stage $k$, meaning $G\notin\Delta_{k-1}$, if and only if
\begin{align*}
G\not\subseteq F_k\setminus\{v\}\text{ for every }v\in A_k.
\end{align*}
For a vertex $v\in F_k$ and a face $G\subseteq F_k$, the condition $G\not\subseteq F_k\setminus\{v\}$ is equivalent to $v\in G$. Hence $G$ is new at stage $k$ if and only if
\begin{align*}
A_k\subseteq G\subseteq F_k.
\end{align*}
Thus the new faces in $F_k$ form the Boolean interval
\begin{align*}
\{G : A_k\subseteq G\subseteq F_k\}.
\end{align*}
By the definition of the restriction face in a shelling, $R(F_k)$ is the unique minimal new face of $F_k$, so $R(F_k)=A_k$. Therefore the new faces at stage $k$ are exactly
\begin{align*}
\{G : R(F_k)\subseteq G\subseteq F_k\}.
\end{align*}
For $k=1$, no earlier facet exists. By convention $R(F_1)=\varnothing$, and every face $G\subseteq F_1$ satisfies
\begin{align*}
R(F_1)\subseteq G\subseteq F_1.
\end{align*}
Thus the same description also holds for $k=1$.
[/step]
[step:Use first appearance to prove the intervals are disjoint and exhaustive]
Let $H\in\Delta$ be any face. Since $F_1,\dots,F_t$ are all the facets of $\Delta$, there exists at least one $k\in\{1,\dots,t\}$ such that $H\subseteq F_k$. Define the first-containing-facet index map
\begin{align*}
m:\Delta&\to\{1,\dots,t\}
\end{align*}
by
\begin{align*}
m(H)=\min\{k\in\{1,\dots,t\}:H\subseteq F_k\}.
\end{align*}
By definition of $m(H)$, the face $H$ is not in $\Delta_{m(H)-1}$, so $H$ is new at stage $m(H)$. From the previous step,
\begin{align*}
R(F_{m(H)})\subseteq H\subseteq F_{m(H)}.
\end{align*}
Thus every face of $\Delta$ lies in at least one restriction interval.
Conversely, if $H$ lies in the interval
\begin{align*}
\{G : R(F_k)\subseteq G\subseteq F_k\},
\end{align*}
then the previous step says that $H$ is new at stage $k$. Hence $H\notin\Delta_{k-1}$, so $H$ is not contained in any earlier facet $F_i$ with $i<k$. Therefore $m(H)=k$.
The interval indexed by $F_k$ is exactly the set of faces $H$ with $m(H)=k$. Since every face has a unique least containing-facet index, these sets are pairwise disjoint and their union is all of $\Delta$. Hence
\begin{align*}
\Delta=\bigsqcup_{k=1}^t \{G\in \Delta : R(F_k)\subseteq G\subseteq F_k\}.
\end{align*}
This includes the empty face: it has first appearance at $F_1$, and $R(F_1)=\varnothing\subseteq\varnothing\subseteq F_1$.
[/step]
