[proofplan]
The proof tracks weights. First, every weight of an irreducible highest weight module lies below its highest weight, and the [tensor product](/page/Tensor%20Product) weight spaces are sums of weight spaces from the two tensor factors. This gives the dominance bound for any highest weight $\nu$ appearing in the tensor product. Then we inspect the top possible weight $\lambda+\mu$: its weight space is the tensor product of the two one-dimensional highest weight spaces, so it is one-dimensional, and the tensor product of highest weight vectors is itself a highest weight vector. Semisimplicity of finite-dimensional complex semisimple [Lie algebra](/page/Lie%20Algebra) modules then gives an irreducible summand $V(\lambda+\mu)$, and the one-dimensional top weight space forces its multiplicity to be exactly $1$.
[/proofplan]
[step:Show that every tensor product weight lies below $\lambda+\mu$]
Let $Q^+\subset\mathfrak h^*$ denote the nonnegative integral span of the positive roots:
\begin{align*}
Q^+:=\left\{\sum_{\alpha\in\Phi^+} n_\alpha\alpha:n_\alpha\in\mathbb Z_{\ge 0}\text{ and }n_\alpha=0\text{ for all but finitely many }\alpha\right\}.
\end{align*}
Since $\Phi^+$ is finite, the finiteness condition is automatic, but it makes the type of the expression explicit.
For a finite-dimensional weight module $M$, write $M_\gamma$ for its $\gamma$-weight space:
\begin{align*}
M_\gamma:=\{m\in M:h\cdot m=\gamma(h)m\text{ for every }h\in\mathfrak h\}.
\end{align*}
Let $\Gamma_\lambda\subset\mathfrak h^*$ denote the set of weights $\gamma$ such that $V(\lambda)_\gamma\neq 0$, and let $\Gamma_\mu\subset\mathfrak h^*$ denote the set of weights $\delta$ such that $V(\mu)_\delta\neq 0$.
By the standard weight bound for finite-dimensional highest weight modules, for every $\gamma\in\Gamma_\lambda$ there exists $\beta_\gamma\in Q^+$ such that
\begin{align*}
\gamma=\lambda-\beta_\gamma.
\end{align*}
Similarly, for every $\delta\in\Gamma_\mu$ there exists $\beta_\delta\in Q^+$ such that
\begin{align*}
\delta=\mu-\beta_\delta.
\end{align*}
By [citetheorem:9387], the tensor product $V(\lambda)\otimes V(\mu)$ is a weight module and its $\eta$-weight space is
\begin{align*}
\bigl(V(\lambda)\otimes V(\mu)\bigr)_\eta
=
\bigoplus_{\gamma+\delta=\eta} V(\lambda)_\gamma\otimes V(\mu)_\delta.
\end{align*}
Therefore, if $\eta$ is a weight of $V(\lambda)\otimes V(\mu)$, then there exist $\gamma\in\Gamma_\lambda$ and $\delta\in\Gamma_\mu$ such that
\begin{align*}
\eta=\gamma+\delta.
\end{align*}
For those weights,
\begin{align*}
\eta
=
(\lambda-\beta_\gamma)+(\mu-\beta_\delta)
=
\lambda+\mu-(\beta_\gamma+\beta_\delta).
\end{align*}
Since $Q^+$ is closed under addition, $\beta_\gamma+\beta_\delta\in Q^+$. Hence every weight $\eta$ of $V(\lambda)\otimes V(\mu)$ satisfies
\begin{align*}
\eta\le \lambda+\mu
\end{align*}
in the dominance order.
[guided]
The point of this step is to turn the tensor product problem into a statement about sums of weights. Define
\begin{align*}
Q^+:=\left\{\sum_{\alpha\in\Phi^+} n_\alpha\alpha:n_\alpha\in\mathbb Z_{\ge 0}\text{ and }n_\alpha=0\text{ for all but finitely many }\alpha\right\}.
\end{align*}
Thus saying that a weight $\eta$ satisfies $\eta\le \lambda+\mu$ means precisely that $\lambda+\mu-\eta\in Q^+$.
For any finite-dimensional weight module $M$, the $\gamma$-weight space is
\begin{align*}
M_\gamma:=\{m\in M:h\cdot m=\gamma(h)m\text{ for every }h\in\mathfrak h\}.
\end{align*}
Let $\Gamma_\lambda$ be the set of weights of $V(\lambda)$ and let $\Gamma_\mu$ be the set of weights of $V(\mu)$. The highest weight property says that all weights of $V(\lambda)$ are obtained from $\lambda$ by subtracting nonnegative combinations of positive roots. Concretely, for every $\gamma\in\Gamma_\lambda$ there is an element $\beta_\gamma\in Q^+$ such that
\begin{align*}
\gamma=\lambda-\beta_\gamma.
\end{align*}
Likewise, for every $\delta\in\Gamma_\mu$ there is an element $\beta_\delta\in Q^+$ such that
\begin{align*}
\delta=\mu-\beta_\delta.
\end{align*}
Now use the tensor product weight decomposition [citetheorem:9387]. Its hypotheses apply because $V(\lambda)$ and $V(\mu)$ are finite-dimensional weight modules over the same Cartan subalgebra $\mathfrak h$. It gives
\begin{align*}
\bigl(V(\lambda)\otimes V(\mu)\bigr)_\eta
=
\bigoplus_{\gamma+\delta=\eta} V(\lambda)_\gamma\otimes V(\mu)_\delta.
\end{align*}
Therefore, a weight $\eta$ of the tensor product can occur only as a sum $\eta=\gamma+\delta$, where $\gamma$ is a weight of $V(\lambda)$ and $\delta$ is a weight of $V(\mu)$. Substituting the highest weight bounds gives
\begin{align*}
\eta
=
(\lambda-\beta_\gamma)+(\mu-\beta_\delta)
=
\lambda+\mu-(\beta_\gamma+\beta_\delta).
\end{align*}
Because $Q^+$ is closed under addition, $\beta_\gamma+\beta_\delta\in Q^+$. Hence $\lambda+\mu-\eta\in Q^+$, which is exactly
\begin{align*}
\eta\le \lambda+\mu.
\end{align*}
[/guided]
[/step]
[step:Apply the weight bound to the highest weight of each irreducible summand]
Suppose $V(\nu)$ occurs as an irreducible direct summand of $V(\lambda)\otimes V(\mu)$. Then there is a $\mathfrak g$-submodule $W\subseteq V(\lambda)\otimes V(\mu)$ such that $W\cong V(\nu)$.
Let $w_\nu\in W$ be a nonzero highest weight vector of weight $\nu$. Since $W$ is a submodule of $V(\lambda)\otimes V(\mu)$, the vector $w_\nu$ is also a vector of weight $\nu$ in $V(\lambda)\otimes V(\mu)$. Therefore $\nu$ is a weight of $V(\lambda)\otimes V(\mu)$. By the previous step,
\begin{align*}
\nu\le \lambda+\mu.
\end{align*}
[/step]
[step:Identify the top weight space as one-dimensional]
We compute the $\lambda+\mu$ weight space of $V(\lambda)\otimes V(\mu)$. By [citetheorem:9387],
\begin{align*}
\bigl(V(\lambda)\otimes V(\mu)\bigr)_{\lambda+\mu}
=
\bigoplus_{\gamma+\delta=\lambda+\mu} V(\lambda)_\gamma\otimes V(\mu)_\delta.
\end{align*}
If $V(\lambda)_\gamma\otimes V(\mu)_\delta\neq 0$ and $\gamma+\delta=\lambda+\mu$, then the first step gives $\gamma\le\lambda$ and $\delta\le\mu$. Hence $\lambda-\gamma\in Q^+$ and $\mu-\delta\in Q^+$. Since
\begin{align*}
(\lambda-\gamma)+(\mu-\delta)=0,
\end{align*}
both terms must be zero in the positive root cone. Thus $\gamma=\lambda$ and $\delta=\mu$.
Consequently,
\begin{align*}
\bigl(V(\lambda)\otimes V(\mu)\bigr)_{\lambda+\mu}
=
V(\lambda)_\lambda\otimes V(\mu)_\mu.
\end{align*}
The highest weight spaces $V(\lambda)_\lambda$ and $V(\mu)_\mu$ are one-dimensional. Therefore
\begin{align*}
\dim \bigl(V(\lambda)\otimes V(\mu)\bigr)_{\lambda+\mu}
=
\dim V(\lambda)_\lambda\cdot \dim V(\mu)_\mu
=
1.
\end{align*}
[/step]
[step:Construct a highest weight vector of weight $\lambda+\mu$]
Choose nonzero highest weight vectors
\begin{align*}
v_\lambda\in V(\lambda)_\lambda
\end{align*}
and
\begin{align*}
v_\mu\in V(\mu)_\mu.
\end{align*}
Define
\begin{align*}
v_{\lambda,\mu}:=v_\lambda\otimes v_\mu\in V(\lambda)\otimes V(\mu).
\end{align*}
For every $h\in\mathfrak h$, the tensor product action gives
\begin{align*}
h\cdot v_{\lambda,\mu}
=
(h\cdot v_\lambda)\otimes v_\mu+v_\lambda\otimes(h\cdot v_\mu)
=
\lambda(h)v_\lambda\otimes v_\mu+\mu(h)v_\lambda\otimes v_\mu
=
(\lambda+\mu)(h)v_{\lambda,\mu}.
\end{align*}
Thus $v_{\lambda,\mu}$ has weight $\lambda+\mu$.
Let $\mathfrak n^+:=\bigoplus_{\alpha\in\Phi^+}\mathfrak g_\alpha$ be the positive nilpotent subalgebra. For every $x\in\mathfrak n^+$, the highest weight property gives
\begin{align*}
x\cdot v_\lambda=0
\end{align*}
and
\begin{align*}
x\cdot v_\mu=0.
\end{align*}
Therefore the tensor product action gives
\begin{align*}
x\cdot v_{\lambda,\mu}
=
(x\cdot v_\lambda)\otimes v_\mu+v_\lambda\otimes(x\cdot v_\mu)
=
0.
\end{align*}
Hence $v_{\lambda,\mu}$ is a highest weight vector of weight $\lambda+\mu$.
[/step]
[step:Use semisimplicity to obtain exactly one copy of $V(\lambda+\mu)$]
Since $\lambda,\mu\in P^+$ and $P^+$ is closed under addition, $\lambda+\mu\in P^+$. Let $M\subseteq V(\lambda)\otimes V(\mu)$ be the $\mathfrak g$-submodule generated by $v_{\lambda,\mu}$. The preceding step shows that $M$ is a highest weight module of highest weight $\lambda+\mu$.
By the standard highest weight classification, $M$ has an irreducible quotient isomorphic to $V(\lambda+\mu)$. Since $V(\lambda)\otimes V(\mu)$ is finite-dimensional, and finite-dimensional modules over a complex semisimple Lie algebra are completely reducible, this quotient occurs as an irreducible direct summand of $V(\lambda)\otimes V(\mu)$. Thus $V(\lambda+\mu)$ occurs.
It remains to determine its multiplicity. Suppose $V(\lambda+\mu)$ occurs with multiplicity $m\in\mathbb Z_{\ge 0}$. Each copy of $V(\lambda+\mu)$ contributes a one-dimensional weight space of weight $\lambda+\mu$, and no irreducible summand with highest weight strictly below $\lambda+\mu$ contributes to that weight space by the dominance bound proved above. Therefore
\begin{align*}
m\le \dim \bigl(V(\lambda)\otimes V(\mu)\bigr)_{\lambda+\mu}=1.
\end{align*}
Since we have already shown that at least one copy occurs, $m\ge 1$. Hence
\begin{align*}
m=1.
\end{align*}
This proves both the dominance bound and the multiplicity-one statement.
[/step]
