[proofplan]
The proof has three ingredients. First, Schur orthogonality gives the exact [inner product](/page/Inner%20Product) of two matrix coefficients, including the normalization factor $d_\pi^{-1}$. Second, the Peter--Weyl density theorem says that the linear span of all finite-dimensional unitary matrix coefficients is dense in $C(G)$, hence dense in $L^2(G,\nu)$. Finally, a direct computation with $c_{\lambda,v}(x)=\lambda(\pi(x)v)$ identifies the $G\times G$ action on each coefficient space with the tensor action on $V_\pi^*\otimes V_\pi$.
[/proofplan]
[step:Declare the matrix coefficient spaces and their Hilbert structures]
Let $C(G)$ denote the complex [vector space](/page/Vector%20Space) of continuous functions $G\to\mathbb C$.
For each $\pi\in\widehat G$, let
\begin{align*}
\mathcal C_\pi:=\operatorname{span}_{\mathbb C}\{c_{\lambda,v}:\lambda\in V_\pi^*,\,v\in V_\pi\}\subset C(G)
\end{align*}
denote the coefficient space of $\pi$.
Choose an [orthonormal basis](/page/Orthonormal%20Basis) $(e_1,\dots,e_{d_\pi})$ of $V_\pi$, and let $(\varepsilon_1,\dots,\varepsilon_{d_\pi})$ be the [dual basis](/theorems/414) of $V_\pi^*$. We equip $V_\pi^*$ with the Hilbert inner product for which this dual basis is orthonormal, and we equip $V_\pi^*\otimes V_\pi$ with the corresponding tensor-product Hilbert inner product. This construction is independent of the chosen orthonormal basis because the dual Hilbert structure is transported by unitary [change of basis](/page/Change%20Of%20Basis).
Define the [linear map](/page/Linear%20Map)
\begin{align*}
T_\pi:V_\pi^*\otimes V_\pi&\to C(G)
\end{align*}
\begin{align*}
\lambda\otimes v&\mapsto c_{\lambda,v}.
\end{align*}
By definition, $\operatorname{im}T_\pi=\mathcal C_\pi$.
[/step]
[step:Use Schur orthogonality to make the normalized coefficient maps unitary]
We use the Schur orthogonality relations for irreducible finite-dimensional unitary representations of compact groups: if $\pi:G\to U(V_\pi)$ and $\sigma:G\to U(V_\sigma)$ are irreducible unitary representations, then for all $\lambda,\mu\in V_\pi^*$ and $v,w\in V_\pi$,
\begin{align*}
\int_G c_{\lambda,v}(x)\overline{c_{\mu,w}(x)}\,d\nu(x)=\frac{1}{d_\pi}(\lambda,\mu)_{V_\pi^*}(v,w)_{V_\pi},
\end{align*}
while if $\pi$ and $\sigma$ are inequivalent irreducible representations, then for all $\lambda\in V_\pi^*$, $v\in V_\pi$, $\alpha\in V_\sigma^*$, and $u\in V_\sigma$,
\begin{align*}
\int_G c_{\lambda,v}(x)\overline{c_{\alpha,u}(x)}\,d\nu(x)=0.
\end{align*}
The inner product on $V_\pi^*$ in this formula is the dual-basis Hilbert structure declared above: if $(e_1,\dots,e_{d_\pi})$ is orthonormal in $V_\pi$ and $(\varepsilon_1,\dots,\varepsilon_{d_\pi})$ is its dual basis, then $(\varepsilon_i,\varepsilon_j)_{V_\pi^*}=\delta_{ij}$. This is the convention under which the displayed Schur orthogonality relation has the stated order of factors. Here this result is being used as a standard prerequisite not yet resolved to a wiki theorem: Schur orthogonality relations.
It follows that
\begin{align*}
(\sqrt{d_\pi}T_\pi(\lambda\otimes v),\sqrt{d_\pi}T_\pi(\mu\otimes w))_{L^2(G,\nu)}=(\lambda,\mu)_{V_\pi^*}(v,w)_{V_\pi}.
\end{align*}
By sesquilinearity, the same identity holds for arbitrary elements of $V_\pi^*\otimes V_\pi$. Thus
\begin{align*}
\sqrt{d_\pi}T_\pi:V_\pi^*\otimes V_\pi\to L^2(G,\nu)
\end{align*}
is an isometric linear embedding. The second Schur orthogonality formula shows that the closed subspaces $\sqrt{d_\pi}\mathcal C_\pi=\mathcal C_\pi$ and $\sqrt{d_\sigma}\mathcal C_\sigma=\mathcal C_\sigma$ are orthogonal whenever $\pi$ and $\sigma$ are inequivalent.
[guided]
The normalization factor $\sqrt{d_\pi}$ is exactly where the Hilbert-space statement becomes precise. Schur orthogonality says that the unnormalized coefficient map loses a factor of $d_\pi^{-1}$ in the $L^2$ inner product:
\begin{align*}
(c_{\lambda,v},c_{\mu,w})_{L^2(G,\nu)}=\frac{1}{d_\pi}(\lambda,\mu)_{V_\pi^*}(v,w)_{V_\pi}.
\end{align*}
Multiplying each coefficient by $\sqrt{d_\pi}$ cancels this factor:
\begin{align*}
(\sqrt{d_\pi}c_{\lambda,v},\sqrt{d_\pi}c_{\mu,w})_{L^2(G,\nu)}=d_\pi\frac{1}{d_\pi}(\lambda,\mu)_{V_\pi^*}(v,w)_{V_\pi}.
\end{align*}
Therefore
\begin{align*}
(\sqrt{d_\pi}c_{\lambda,v},\sqrt{d_\pi}c_{\mu,w})_{L^2(G,\nu)}=(\lambda,\mu)_{V_\pi^*}(v,w)_{V_\pi}.
\end{align*}
This is precisely the tensor-product Hilbert inner product on simple tensors:
\begin{align*}
(\lambda\otimes v,\mu\otimes w)_{V_\pi^*\otimes V_\pi}=(\lambda,\mu)_{V_\pi^*}(v,w)_{V_\pi}.
\end{align*}
Since simple tensors span $V_\pi^*\otimes V_\pi$, sesquilinearity extends the identity to all tensors. Hence the map
\begin{align*}
\sqrt{d_\pi}T_\pi:V_\pi^*\otimes V_\pi\to L^2(G,\nu)
\end{align*}
preserves inner products and is therefore injective.
Schur orthogonality also separates different irreducible representations. If $\pi$ and $\sigma$ are inequivalent irreducibles, then every coefficient of $\pi$ is orthogonal in $L^2(G,\nu)$ to every coefficient of $\sigma$:
\begin{align*}
(c_{\lambda,v},c_{\alpha,u})_{L^2(G,\nu)}=0.
\end{align*}
Thus the coefficient spaces belonging to distinct isomorphism classes form mutually orthogonal finite-dimensional subspaces of $L^2(G,\nu)$.
[/guided]
[/step]
[step:Use Peter--Weyl density to obtain the whole Hilbert space]
Let
\begin{align*}
\mathcal A:=\operatorname{span}_{\mathbb C}\{c_{\lambda,v}:\pi\in\widehat G,\,\lambda\in V_\pi^*,\,v\in V_\pi\}\subset C(G)
\end{align*}
be the algebraic span of all irreducible matrix coefficients.
By the Peter--Weyl density theorem for compact groups in its uniform-density form, the subspace $\mathcal A$ is uniformly dense in $C(G)$. This density statement is the algebraic uniform-density prerequisite, not the Hilbert-space direct-sum conclusion being proved here. Since $G$ is compact and $\nu(G)=1$, [uniform convergence](/page/Uniform%20Convergence) implies $L^2(G,\nu)$ convergence. Indeed, for $f\in C(G)$,
\begin{align*}
\|f\|_{L^2(G,\nu)}^2=\int_G |f(x)|^2\,d\nu(x)\leq \|f\|_\infty^2\nu(G)=\|f\|_\infty^2.
\end{align*}
Thus $\mathcal A$ is dense in the closure of $C(G)$ in $L^2(G,\nu)$. Since continuous functions are dense in $L^2(G,\nu)$ for the normalized Haar measure on the compact Lie group $G$, the subspace $\mathcal A$ is dense in $L^2(G,\nu)$.
Combining this density with the mutual orthogonality of the spaces $\mathcal C_\pi$, we obtain
\begin{align*}
L^2(G,\nu)=\widehat{\bigoplus}_{\pi\in\widehat G}\mathcal C_\pi.
\end{align*}
Using the unitary identifications $\sqrt{d_\pi}T_\pi:V_\pi^*\otimes V_\pi\to\mathcal C_\pi$, this becomes the [Hilbert space](/page/Hilbert%20Space) decomposition
\begin{align*}
L^2(G,\nu)\cong\widehat{\bigoplus}_{\pi\in\widehat G}(V_\pi^*\otimes V_\pi).
\end{align*}
[/step]
[step:Compute the $G\times G$ action on matrix coefficients]
Define the action of $G\times G$ on $L^2(G,\nu)$ by
\begin{align*}
((g,h)\cdot f)(x)=f(g^{-1}xh)
\end{align*}
for $g,h,x\in G$ and $f\in L^2(G,\nu)$. This action is unitary because Haar measure on compact groups is both left and right invariant; right invariance may be obtained, for compact groups, from unimodularity, for example from [[Compact Lie Groups Are Unimodular](/theorems/8829)][citetheorem:8829].
Fix $\pi\in\widehat G$, $\lambda\in V_\pi^*$, $v\in V_\pi$, and $g,h,x\in G$. Then
\begin{align*}
((g,h)\cdot c_{\lambda,v})(x)=c_{\lambda,v}(g^{-1}xh).
\end{align*}
By the definition of $c_{\lambda,v}$,
\begin{align*}
c_{\lambda,v}(g^{-1}xh)=\lambda(\pi(g^{-1}xh)v).
\end{align*}
Since $\pi$ is a [group homomorphism](/page/Group%20Homomorphism),
\begin{align*}
\lambda(\pi(g^{-1}xh)v)=\lambda(\pi(g^{-1})\pi(x)\pi(h)v).
\end{align*}
Let $GL(V_\pi^*)$ denote the group of invertible complex-linear maps $V_\pi^*\to V_\pi^*$. Define the dual representation
\begin{align*}
\pi^*:G&\to GL(V_\pi^*)
\end{align*}
\begin{align*}
g&\mapsto \pi^*(g),
\end{align*}
where
\begin{align*}
\pi^*(g)\lambda=\lambda\circ\pi(g^{-1}).
\end{align*}
Then the preceding equality becomes
\begin{align*}
((g,h)\cdot c_{\lambda,v})(x)=c_{\pi^*(g)\lambda,\pi(h)v}(x).
\end{align*}
Thus the action on the coefficient space corresponding to $\pi$ is exactly the tensor action
\begin{align*}
(g,h)\cdot(\lambda\otimes v)=(\pi^*(g)\lambda)\otimes(\pi(h)v).
\end{align*}
Because the scalar factor $\sqrt{d_\pi}$ is constant on the summand, the same action formula holds under the unitary map $\sqrt{d_\pi}T_\pi$.
[/step]
[step:Extract the orthonormal basis formulation]
For each $\pi\in\widehat G$, choose an orthonormal basis $(e_1,\dots,e_{d_\pi})$ of $V_\pi$ and its dual orthonormal basis $(\varepsilon_1,\dots,\varepsilon_{d_\pi})$ of $V_\pi^*$. The tensors
\begin{align*}
\varepsilon_i\otimes e_j
\end{align*}
with $1\leq i,j\leq d_\pi$ form an orthonormal basis of $V_\pi^*\otimes V_\pi$. Since $\sqrt{d_\pi}T_\pi$ is unitary onto $\mathcal C_\pi$, the functions
\begin{align*}
\sqrt{d_\pi}\,c_{\varepsilon_i,e_j}
\end{align*}
form an orthonormal basis of $\mathcal C_\pi$. Taking the union over all $\pi\in\widehat G$ gives an orthonormal basis of the Hilbert direct sum
\begin{align*}
\widehat{\bigoplus}_{\pi\in\widehat G}\mathcal C_\pi=L^2(G,\nu).
\end{align*}
This proves the stated Hilbert space decomposition, the $G\times G$-equivariance, and the equivalent orthonormal basis formulation.
[/step]
