[proofplan]
We separate a tableau in $B_m(\nu)$ into the boxes containing the maximal letter $m$ and the remaining boxes. The restricted operators $e_i,f_i$ for $1\le i\le m-2$ only see the letters $i$ and $i+1$, so they preserve the set of boxes containing $m$. Semistandardness forces those $m$-boxes to form a horizontal strip, and deleting them gives a partition $\mu$ interlacing $\nu$. Conversely, every interlacing $\mu$ gives an embedded copy of $B_{m-1}(\mu)$ by filling the skew horizontal strip $\nu/\mu$ with $m$.
[/proofplan]
[step:Restrict the crystal operators and isolate the boxes containing $m$]
Let $Y(\nu)$ denote the Young diagram of $\nu$, written as
\begin{align*}
Y(\nu)=\{(a,c):1\le a\le m,\ 1\le c\le \nu_a\}.
\end{align*}
Define the $m$-box map as the function $D_m:B_m(\nu)\to \{D:D\subset Y(\nu)\}$. Thus the codomain is the set of all subsets of the Young diagram $Y(\nu)$. For a tableau $T\in B_m(\nu)$, set
\begin{align*}
D_m(T):=\{(a,c)\in Y(\nu):T(a,c)=m\}.
\end{align*}
For each $1\le i\le m-2$, the Kashiwara operators $e_i$ and $f_i$ on semistandard tableaux are defined from the $i$-signature, which is obtained from the reading word by retaining only the letters $i$ and $i+1$; this is the semistandard tableau model for type $A$ highest weight crystals, [citetheorem:8475]. Since $m$ is neither $i$ nor $i+1$, applying $e_i$ or $f_i$ never changes a box with entry $m$ and never changes another entry into $m$. Hence, whenever $e_iT\ne 0$ or $f_iT\ne 0$, we have
\begin{align*}
D_m(e_iT)=D_m(T).
\end{align*}
Likewise,
\begin{align*}
D_m(f_iT)=D_m(T).
\end{align*}
Thus every connected component of the restricted $\mathfrak{gl}_{m-1}$-crystal is contained in the set of tableaux with a fixed subset $D\subseteq Y(\nu)$ of $m$-boxes.
[/step]
[step:Show that deleting the $m$-boxes gives an interlacing partition]
Fix $T\in B_m(\nu)$, and write $D=D_m(T)$. Since rows of $T$ are weakly increasing and $m$ is the largest allowed entry, the boxes of $D$ in row $a$ form a right-justified suffix of that row. Define integers $\mu_a$ for $1\le a\le m$ by
\begin{align*}
\mu_a:=|\{c:1\le c\le \nu_a,\ T(a,c)<m\}|.
\end{align*}
Then $0\le \mu_a\le \nu_a$, and the boxes not in $D$ are exactly the row-initial set
\begin{align*}
\{(a,c):1\le a\le m,\ 1\le c\le \mu_a\}.
\end{align*}
We prove that $\mu=(\mu_1,\dots,\mu_m)$ is a partition with $\mu_m=0$ and satisfies $\nu_a\ge \mu_a\ge \nu_{a+1}$ for $1\le a\le m-1$; after this verification, the displayed row-initial set is the Young diagram $Y(\mu)$. The inequality $\mu_a\le \nu_a$ follows from the definition. For the lower inequality, suppose $1\le a\le m-1$ and $1\le c\le \nu_{a+1}$. If $(a,c)$ were an $m$-box, then the box $(a+1,c)$ would exist below it and semistandard column-strictness would give
\begin{align*}
m=T(a,c)<T(a+1,c)\le m,
\end{align*}
a contradiction. Therefore every box $(a,c)$ with $1\le c\le \nu_{a+1}$ is not an $m$-box, so $\mu_a\ge \nu_{a+1}$.
It remains to prove that no non-$m$ box survives in row $m$. If $1\le c\le \nu_m$, then the boxes $(1,c),(2,c),\dots,(m,c)$ all lie in $Y(\nu)$. Column-strictness gives
\begin{align*}
T(1,c)<T(2,c)<\cdots<T(m,c),
\end{align*}
and these are $m$ strictly increasing elements of the $m$-element set $\{1,\dots,m\}$. Hence they must be $1,2,\dots,m$ in order, so $T(m,c)=m$. Thus every box in row $m$ is an $m$-box, and therefore $\mu_m=0$.
Thus deleting the $m$-boxes gives a partition $\mu=(\mu_1,\dots,\mu_{m-1})$ satisfying
\begin{align*}
\nu_a\ge \mu_a\ge \nu_{a+1}
\end{align*}
for every $1\le a\le m-1$. This is exactly the condition that $\nu/\mu$ contains at most one box in each column, hence that $\nu/\mu$ is a horizontal strip.
[guided]
The point of this step is to turn the positions of the letter $m$ into a shape. Define the $m$-box map as the function $D_m:B_m(\nu)\to \{D:D\subset Y(\nu)\}$. Here the codomain is the set of all subsets of the Young diagram $Y(\nu)$. For $T\in B_m(\nu)$, set
\begin{align*}
D_m(T):=\{(a,c)\in Y(\nu):T(a,c)=m\}.
\end{align*}
Because each row of a semistandard tableau is weakly increasing and $m$ is the largest letter in the alphabet $\{1,\dots,m\}$, once an $m$ appears in a row, every box to its right is also an $m$. Therefore the non-$m$ boxes in row $a$ form an initial segment of that row. Define
\begin{align*}
\mu_a:=|\{c:1\le c\le \nu_a,\ T(a,c)<m\}|
\end{align*}
for $1\le a\le m$. Then the non-$m$ boxes are precisely the row-initial set
\begin{align*}
\{(a,c):1\le a\le m,\ 1\le c\le \mu_a\}.
\end{align*}
Once we prove that $\mu$ is a partition, this set will be the Young diagram $Y(\mu)$.
We now verify the interlacing inequalities. The upper bound $\mu_a\le \nu_a$ is part of the definition: the non-$m$ boxes in row $a$ are a subset of all boxes in row $a$. For the lower bound, fix $a$ with $1\le a\le m-1$. We claim that the first $\nu_{a+1}$ boxes in row $a$ cannot contain $m$. Indeed, if $1\le c\le \nu_{a+1}$ and $T(a,c)=m$, then the box $(a+1,c)$ exists below it. Since $T$ is semistandard, columns are strictly increasing from top to bottom, so
\begin{align*}
T(a,c)<T(a+1,c).
\end{align*}
Substituting $T(a,c)=m$ gives
\begin{align*}
m<T(a+1,c).
\end{align*}
This is impossible because all entries of $T$ lie in $\{1,\dots,m\}$. Hence every box $(a,c)$ with $1\le c\le \nu_{a+1}$ is a non-$m$ box, and therefore
\begin{align*}
\mu_a\ge \nu_{a+1}.
\end{align*}
Combining this with $\mu_a\le \nu_a$, we obtain
\begin{align*}
\nu_a\ge \mu_a\ge \nu_{a+1}
\end{align*}
for every $1\le a\le m-1$.
We also have to check that the remaining shape has no $m$th row, because a $\mathfrak{gl}_{m-1}$ highest weight component is indexed by a partition with at most $m-1$ parts. Let $c$ be an integer with $1\le c\le \nu_m$. Since $\nu$ has $m$ rows of length at least $c$, the boxes
\begin{align*}
(1,c),(2,c),\dots,(m,c)
\end{align*}
all lie in $Y(\nu)$. Column-strictness gives a strictly increasing chain
\begin{align*}
T(1,c)<T(2,c)<\cdots<T(m,c)
\end{align*}
of $m$ entries from the alphabet $\{1,\dots,m\}$. The only strictly increasing sequence of length $m$ in this alphabet is $1,2,\dots,m$, so the bottom entry satisfies $T(m,c)=m$. Hence every box in row $m$ is deleted, which proves $\mu_m=0$.
These inequalities are the usual interlacing inequalities. They are also equivalent to saying that $\nu/\mu$ has at most one box in each column: if two skew boxes occurred in the same column in consecutive rows, then the lower bound $\mu_a\ge \nu_{a+1}$ would fail for the row above.
Finally, the filling left on $Y(\mu)$ is semistandard. It uses only entries in $\{1,\dots,m-1\}$ because all boxes with entry $m$ were deleted. Its rows remain weakly increasing and its columns remain strictly increasing because those inequalities are inherited by restricting the original semistandard tableau $T$ to the subdiagram $Y(\mu)$.
[/guided]
[/step]
[step:Embed each interlacing $\mu$ as a restricted subcrystal]
Conversely, let $\mu=(\mu_1,\dots,\mu_{m-1})$ be a partition satisfying
\begin{align*}
\nu_a\ge \mu_a\ge \nu_{a+1}
\end{align*}
for every $1\le a\le m-1$, and set $\mu_m:=0$. Define
\begin{align*}
\Phi_\mu:B_{m-1}(\mu)\to B_m(\nu)
\end{align*}
as follows. For $U\in B_{m-1}(\mu)$, let $\Phi_\mu(U)$ be the tableau of shape $\nu$ obtained by placing the entry $U(a,c)$ in each box $(a,c)\in Y(\mu)$ and placing the entry $m$ in each box of $Y(\nu)\setminus Y(\mu)$.
We check that $\Phi_\mu(U)$ is semistandard. Along rows, $U$ is weakly increasing on $Y(\mu)$, and all added boxes lie to the right of $Y(\mu)$ in their rows and have entry $m$, the largest possible entry. Along columns, pairs of boxes both lying in $Y(\mu)$ are strictly increasing because $U$ is semistandard. Now let $(a,c)\in Y(\nu)\setminus Y(\mu)$ be an added $m$-box. If it has an upper neighbour, then $a\ge 2$ and $c\le \nu_a$. The interlacing inequality in row $a-1$ gives $\mu_{a-1}\ge \nu_a$, so $c\le\mu_{a-1}$ and the upper neighbour $(a-1,c)$ lies in $Y(\mu)$; its entry is therefore in $\{1,\dots,m-1\}$ and is strictly less than $m$. If the added box $(a,c)$ had a lower neighbour in $Y(\nu)$, then $c>\mu_a$ and $c\le \nu_{a+1}$, contradicting $\mu_a\ge \nu_{a+1}$. Thus no added $m$-box has a lower neighbour, and column-strictness holds.
Therefore $\Phi_\mu$ is well-defined. It is injective because restricting $\Phi_\mu(U)$ to $Y(\mu)$ recovers $U$.
[/step]
[step:Verify that the embedding commutes with the restricted crystal operators]
Let $1\le i\le m-2$. For $U\in B_{m-1}(\mu)$, the $i$-signature of $U$ is computed using only the letters $i$ and $i+1$. The tableau $\Phi_\mu(U)$ has exactly the same letters $i$ and $i+1$ in exactly the same relative reading order, because the added boxes contain only the letter $m$. Hence the uncancelled signs in the $i$-signature are the same for $U$ and for $\Phi_\mu(U)$.
Extend $\Phi_\mu$ to the formal null symbol by setting $\Phi_\mu(0):=0$. It follows directly from the definition of the Kashiwara operators on semistandard tableaux that
\begin{align*}
\Phi_\mu(e_iU)=e_i\Phi_\mu(U)
\end{align*}
whenever $e_iU\ne 0$, and both sides are $0$ when $e_iU=0$. Similarly,
\begin{align*}
\Phi_\mu(f_iU)=f_i\Phi_\mu(U)
\end{align*}
whenever $f_iU\ne 0$, and both sides are $0$ when $f_iU=0$. Thus $\Phi_\mu$ is an isomorphism from $B_{m-1}(\mu)$ onto its image as a restricted $\mathfrak{gl}_{m-1}$-crystal.
[/step]
[step:Identify the connected components and rule out multiplicities]
By the semistandard tableau model for type $A$ highest weight crystals, [citetheorem:8475], $B_{m-1}(\mu)$ is the connected highest weight crystal of highest weight $\mu$. Since $\Phi_\mu$ commutes with all restricted operators $e_i,f_i$ for $1\le i\le m-2$, its image is a connected restricted subcrystal isomorphic to $B_{m-1}(\mu)$.
Every tableau $T\in B_m(\nu)$ belongs to exactly one such image: delete its $m$-boxes to obtain the partition $\mu$ constructed above. The remaining filling has entries in $\{1,\dots,m-1\}$ because all boxes with entry $m$ were deleted, and it is semistandard on $Y(\mu)$ because the row weak increase and column strict increase are inherited by restricting the original semistandard tableau $T$ to the subdiagram $Y(\mu)$. Hence the remaining filling is an element of $B_{m-1}(\mu)$, and its image under $\Phi_\mu$ is $T$. Conversely, different partitions $\mu$ give different subsets $Y(\nu)\setminus Y(\mu)$ of $m$-boxes, so their images are disjoint. Hence the restricted crystal decomposes as the disjoint union
\begin{align*}
\operatorname{Res}^{\mathfrak{gl}_m}_{\mathfrak{gl}_{m-1}}B_m(\nu)
\cong
\bigsqcup_{\mu:\ \nu_a\ge \mu_a\ge \nu_{a+1}} B_{m-1}(\mu),
\end{align*}
where $1\le a\le m-1$ in the displayed condition.
The inequalities $\nu_a\ge \mu_a\ge \nu_{a+1}$ are equivalent to $\nu/\mu$ being a horizontal strip, so this is precisely the asserted semistandard branching rule.
[/step]
