[proofplan]
We identify the crystal of $V\otimes W$ with the [tensor product](/page/Tensor%20Product) crystal $B(V)\otimes B(W)$ using the tensor product theorem for crystal bases. Since the category is semisimple, $V\otimes W$ is a direct sum of irreducible highest-weight modules, and the crystal of such a direct sum is the disjoint union of the corresponding highest-weight crystals. Each copy of $B(\lambda)$ has exactly one highest-weight vertex of weight $\lambda$, so counting highest-weight vertices in the tensor product crystal gives exactly the multiplicity of $V(\lambda)$.
[/proofplan]
[step:Identify the tensor product crystal with the crystal of $V\otimes W$]
Let $I$ denote the set of simple-root indices for the type $A$ root datum. Let
\begin{align*}
e_i,f_i:B(V)\to B(V)\sqcup\{0\}
\end{align*}
and
\begin{align*}
e_i,f_i:B(W)\to B(W)\sqcup\{0\}
\end{align*}
denote the Kashiwara operators for $i\in I$.
By the tensor-product compatibility assumed in the statement, applied to the objects $V$ and $W$, the tensor product module $V\otimes W$ has crystal
\begin{align*}
B(V\otimes W)\cong B(V)\otimes B(W),
\end{align*}
where the operators on the right are the tensor product crystal operators. Thus it is enough to prove the claim for the crystal $B(V\otimes W)$.
[/step]
[step:Decompose the crystal of the semisimple tensor product]
By hypothesis, $V\otimes W$ decomposes in $\mathcal C$ as
\begin{align*}
V\otimes W \cong \bigoplus_{\lambda} V(\lambda)^{\oplus m_\lambda}.
\end{align*}
For each dominant weight $\lambda$, let $B(\lambda)$ denote the normal highest-weight crystal of the irreducible highest-weight module $V(\lambda)$, and let $u_\lambda\in B(\lambda)$ denote its unique highest-weight vertex.
By the direct-sum compatibility assumed in the statement, the crystal of a finite direct sum is the disjoint union of the crystals of its summands. Hence there is an isomorphism of crystals
\begin{align*}
B(V\otimes W)\cong \bigsqcup_{\lambda}\bigsqcup_{a=1}^{m_\lambda} B(\lambda,a),
\end{align*}
where $B(\lambda,a)$ is a copy of $B(\lambda)$ for each integer $a$ satisfying $1\le a\le m_\lambda$. Each copy $B(\lambda,a)$ is connected and highest-weight, with unique highest-weight vertex corresponding to $u_\lambda$.
[guided]
The semisimple decomposition is not merely a vector-space decomposition; it is a decomposition in the highest-weight module category:
\begin{align*}
V\otimes W \cong \bigoplus_{\lambda} V(\lambda)^{\oplus m_\lambda}.
\end{align*}
Here $V(\lambda)$ is the irreducible module with highest weight $\lambda$, and $m_\lambda$ records how many copies of that irreducible appear.
The statement assumes the needed finite direct-sum compatibility: the crystal of a finite direct sum is the disjoint union of the crystals of the summands. Applying that compatibility to the displayed semisimple decomposition gives
\begin{align*}
B(V\otimes W)\cong \bigsqcup_{\lambda}\bigsqcup_{a=1}^{m_\lambda} B(\lambda,a),
\end{align*}
where $B(\lambda,a)$ denotes the copy of $B(\lambda)$ coming from the $a$th summand isomorphic to $V(\lambda)$.
By the first hypothesis in the statement, each irreducible highest-weight module $V(\lambda)$ has a connected normal highest-weight crystal $B(\lambda)$ with a unique highest-weight vertex $u_\lambda$. Its defining properties are
\begin{align*}
\operatorname{wt}(u_\lambda)=\lambda
\end{align*}
and
\begin{align*}
e_i u_\lambda=0
\end{align*}
for every $i\in I$. Hence each summand copy $B(\lambda,a)$ contributes exactly one highest-weight vertex of weight $\lambda$.
[/guided]
[/step]
[step:Identify the connected components]
The disjoint union decomposition
\begin{align*}
B(V\otimes W)\cong \bigsqcup_{\lambda}\bigsqcup_{a=1}^{m_\lambda} B(\lambda,a)
\end{align*}
is a decomposition by connected crystals, because each $B(\lambda,a)$ is connected and the Kashiwara operators preserve each summand copy. Therefore every connected component of $B(V\otimes W)$ is one of the highest-weight crystals $B(\lambda,a)$.
Using the isomorphism
\begin{align*}
B(V\otimes W)\cong B(V)\otimes B(W),
\end{align*}
the same conclusion holds for the connected components of $B(V)\otimes B(W)$.
[/step]
[step:Count highest-weight vertices of each weight]
Fix a dominant weight $\lambda$. In the decomposition
\begin{align*}
B(V\otimes W)\cong \bigsqcup_{\mu}\bigsqcup_{a=1}^{m_\mu} B(\mu,a),
\end{align*}
a component $B(\mu,a)$ has a highest-weight vertex of weight $\mu$, and this vertex is unique. Thus the components contributing highest-weight vertices of weight $\lambda$ are exactly
\begin{align*}
B(\lambda,1),B(\lambda,2),\dots,B(\lambda,m_\lambda).
\end{align*}
Hence the number of vertices $b\in B(V\otimes W)$ satisfying
\begin{align*}
\operatorname{wt}(b)=\lambda
\end{align*}
and
\begin{align*}
e_i b=0
\end{align*}
for every $i\in I$ is $m_\lambda$.
Transporting this count through the crystal isomorphism
\begin{align*}
B(V\otimes W)\cong B(V)\otimes B(W)
\end{align*}
proves that the number of highest-weight vertices of weight $\lambda$ in $B(V)\otimes B(W)$ is exactly the multiplicity of $V(\lambda)$ in $V\otimes W$. This proves both assertions.
[/step]
