[proofplan]
We prove the stronger statement that every abelian subgroup of $U(n)$ admits an [orthonormal basis](/page/Orthonormal%20Basis) of simultaneous eigenvectors. The proof is by induction on $n$. If all elements are scalar, any orthonormal basis works; otherwise we choose a non-scalar element, use the spectral theorem for unitary matrices to decompose $\mathbb C^n$ into its orthogonal eigenspaces, observe that commutativity forces every element of $A$ to preserve each eigenspace, and apply induction on those smaller eigenspaces. Concatenating the resulting orthonormal bases gives a unitary [change of basis](/page/Change%20Of%20Basis) that diagonalizes every element of $A$ at once.
[/proofplan]
[step:Prove the simultaneous eigenbasis statement by induction on the dimension]
Throughout this proof, $\mathbb N=\{1,2,3,\dots\}$. For each $n\in\mathbb N$, let $I_n:\mathbb C^n\to\mathbb C^n$ denote the identity [linear map](/page/Linear%20Map), and let $P(n)$ denote the assertion that every abelian subgroup $A\le U(n)$ admits an orthonormal basis $(b_1,\dots,b_n)$ of $\mathbb C^n$ such that every $g\in A$ satisfies
\begin{align*}
g b_i=\lambda_i(g)b_i
\end{align*}
for some scalar $\lambda_i(g)\in U(1)$ and every $1\le i\le n$.
For $n=1$, the standard basis vector $e_1\in\mathbb C$ is an eigenvector for every element of $U(1)$, so $P(1)$ holds.
Assume now that $n\ge 2$ and that $P(m)$ holds for every integer $m$ with $1\le m<n$. Let $A\le U(n)$ be abelian. We work in $\mathbb C^n$ with its standard Hermitian [inner product](/page/Inner%20Product) $(\cdot,\cdot)_{\mathbb C^n}$. Let $I_n:\mathbb C^n\to\mathbb C^n$ denote the identity linear map.
[/step]
[step:Handle the case where every element of $A$ is scalar]
Suppose first that every $g\in A$ is scalar, so for each $g\in A$ there exists $\lambda(g)\in U(1)$ such that
\begin{align*}
g=\lambda(g)I_n.
\end{align*}
Then the standard orthonormal basis $(e_1,\dots,e_n)$ of $\mathbb C^n$ is a simultaneous eigenbasis for $A$, since
\begin{align*}
g e_i=\lambda(g)e_i
\end{align*}
for every $g\in A$ and every $1\le i\le n$. Hence $P(n)$ holds in this case.
[/step]
[step:Decompose by the eigenspaces of a non-scalar element]
It remains to consider the case where some element of $A$ is not scalar. Choose $a\in A$ such that $a\ne \lambda I_n$ for every $\lambda\in\mathbb C$. Since $a\in U(n)$, the spectral theorem for unitary matrices gives distinct complex numbers $\alpha_1,\dots,\alpha_r\in U(1)$ and non-zero subspaces $E_1,\dots,E_r\le\mathbb C^n$ such that
\begin{align*}
E_j=\ker(a-\alpha_j I_n)
\end{align*}
for each $1\le j\le r$, and
\begin{align*}
\mathbb C^n=E_1\oplus\cdots\oplus E_r
\end{align*}
as an orthogonal [direct sum](/page/Direct%20Sum). Here we use the spectral theorem for unitary matrices.
Because $a$ is not scalar, we have $r\ge 2$. Hence each eigenspace has dimension strictly smaller than $n$:
\begin{align*}
1\le \dim E_j<n
\end{align*}
for every $1\le j\le r$.
[/step]
[step:Show every element of $A$ preserves each eigenspace of the chosen element]
Fix $g\in A$ and fix $1\le j\le r$. Since $A$ is abelian, $ga=ag$. If $v\in E_j$, then $av=\alpha_j v$, and therefore
\begin{align*}
a(gv)=g(av)=g(\alpha_j v)=\alpha_j gv.
\end{align*}
Thus $gv\in E_j$. Therefore
\begin{align*}
g(E_j)\subset E_j
\end{align*}
for every $g\in A$ and every $1\le j\le r$.
Since $g\in U(n)$ is invertible and $g^{-1}\in A$, the same argument applied to $g^{-1}$ gives
\begin{align*}
g^{-1}(E_j)\subset E_j.
\end{align*}
Applying $g$ to this inclusion gives $E_j\subset g(E_j)$. Hence
\begin{align*}
g(E_j)=E_j.
\end{align*}
[guided]
Fix an element $g\in A$ and an eigenspace $E_j=\ker(a-\alpha_j I_n)$ of the chosen non-scalar element $a$. We want to prove that $g$ restricts to a unitary operator on $E_j$, so the first point is to show that vectors in $E_j$ are sent back into $E_j$.
Let $v\in E_j$. By definition of $E_j$, this means
\begin{align*}
av=\alpha_j v.
\end{align*}
Because $A$ is abelian and both $a$ and $g$ lie in $A$, we have $ag=ga$. Therefore
\begin{align*}
a(gv)=g(av)=g(\alpha_j v)=\alpha_j gv.
\end{align*}
This equation says exactly that $gv$ is again an eigenvector of $a$ with eigenvalue $\alpha_j$, or possibly the zero vector, which is also contained in the eigenspace. Hence
\begin{align*}
g(E_j)\subset E_j.
\end{align*}
We also need equality, not only inclusion, because the induction hypothesis will be applied to a group of unitary operators on $E_j$. Since $g\in U(n)$, it is invertible, and since $A$ is a subgroup, $g^{-1}\in A$. Repeating the same argument with $g^{-1}$ in place of $g$ gives
\begin{align*}
g^{-1}(E_j)\subset E_j.
\end{align*}
Now apply $g$ to both sides of this inclusion. The left-hand side becomes $E_j$, and the right-hand side becomes $g(E_j)$, so
\begin{align*}
E_j\subset g(E_j).
\end{align*}
Together with $g(E_j)\subset E_j$, this proves
\begin{align*}
g(E_j)=E_j.
\end{align*}
Thus every element of $A$ preserves every eigenspace of $a$.
[/guided]
[/step]
[step:Apply induction on each preserved eigenspace]
For each $1\le j\le r$, define
\begin{align*}
A_j:=\{g|_{E_j}:E_j\to E_j\mid g\in A\}.
\end{align*}
Because each $g\in A$ preserves $E_j$, this is a well-defined set of linear maps on $E_j$. Since $g\in U(n)$ preserves the standard Hermitian inner product on $\mathbb C^n$ and satisfies $g(E_j)=E_j$, the restriction $g|_{E_j}:E_j\to E_j$ preserves the inherited Hermitian inner product on $E_j$. Thus every element of $A_j$ is unitary on $E_j$. If $g,h\in A$, then
\begin{align*}
(g|_{E_j})(h|_{E_j})=(gh)|_{E_j},
\end{align*}
and $gh\in A$ because $A$ is a subgroup, so $A_j$ is closed under composition. The identity element of $A_j$ is $I_n|_{E_j}:E_j\to E_j$, which is the identity map on $E_j$, because $I_n\in A$. If $g\in A$, then $g^{-1}\in A$ and $g^{-1}(E_j)=E_j$, so the inverse of $g|_{E_j}:E_j\to E_j$ is $g^{-1}|_{E_j}:E_j\to E_j$. Therefore $A_j$ is a subgroup of the unitary group $U(E_j)$ of the Hermitian space $E_j$. It is abelian because $A$ is abelian.
Let $m_j:=\dim E_j$. Choose a unitary isomorphism $\Phi_j:\mathbb C^{m_j}\to E_j$. Conjugating by $\Phi_j$ identifies $A_j\le U(E_j)$ with the abelian subgroup
\begin{align*}
\Phi_j^{-1}A_j\Phi_j\le U(m_j).
\end{align*}
We have $1\le m_j<n$, so the induction hypothesis applies to $\Phi_j^{-1}A_j\Phi_j\le U(m_j)$. Transporting the resulting basis through $\Phi_j$, there exists an orthonormal basis
\begin{align*}
(b_{j,1},\dots,b_{j,m_j})
\end{align*}
of $E_j$ such that every operator $g|_{E_j}$ with $g\in A$ is diagonal in this basis. Equivalently, for every $g\in A$ and every $1\le k\le m_j$, there exists $\lambda_{j,k}(g)\in U(1)$ such that
\begin{align*}
g b_{j,k}=\lambda_{j,k}(g)b_{j,k}.
\end{align*}
[/step]
[step:Concatenate the eigenspace bases to obtain a global unitary diagonalization]
Because
\begin{align*}
\mathbb C^n=E_1\oplus\cdots\oplus E_r
\end{align*}
is an orthogonal direct sum, concatenating the orthonormal bases of the spaces $E_j$ gives an orthonormal basis
\begin{align*}
\mathcal B=(b_{1,1},\dots,b_{1,m_1},b_{2,1},\dots,b_{r,m_r})
\end{align*}
of $\mathbb C^n$. The previous step shows that every vector in $\mathcal B$ is an eigenvector for every element of $A$. Hence $P(n)$ holds. By induction, $P(n)$ holds for every $n\in\mathbb N$.
[/step]
[step:Translate the simultaneous eigenbasis into conjugation into the diagonal torus]
Apply $P(n)$ to the given subgroup $A\le U(n)$, and let
\begin{align*}
\mathcal B=(b_1,\dots,b_n)
\end{align*}
be an orthonormal simultaneous eigenbasis of $\mathbb C^n$ for $A$. Define the linear map $q:\mathbb C^n\to\mathbb C^n$ by declaring
\begin{align*}
q e_i=b_i
\end{align*}
for every standard basis vector $e_i$ with $1\le i\le n$, and extending linearly. Since $q$ sends the standard orthonormal basis to the orthonormal basis $\mathcal B$, we have $q\in U(n)$.
For each $g\in A$ and each $1\le i\le n$, choose the scalar $\lambda_i(g)\in U(1)$ such that
\begin{align*}
g b_i=\lambda_i(g)b_i.
\end{align*}
Then
\begin{align*}
q^{-1}gq e_i=q^{-1}g b_i=q^{-1}(\lambda_i(g)b_i)=\lambda_i(g)e_i.
\end{align*}
Thus
\begin{align*}
q^{-1}gq=\operatorname{diag}(\lambda_1(g),\dots,\lambda_n(g))\in T_{U(n)}
\end{align*}
for every $g\in A$. Setting $u:=q^{-1}\in U(n)$ gives
\begin{align*}
uAu^{-1}\subset T_{U(n)}.
\end{align*}
This is the desired conclusion.
[/step]
