[proofplan]
The argument is the standard Hilbert-space orthogonality argument for a self-adjoint operator. First we use self-adjointness to show that the two eigenvalues are real, because each eigenform has positive Petersson norm. Then we compute $(T_n f,g)_{\mathrm{Pet}}$ in two ways: once from the eigenvalue equation for $f$, and once from self-adjointness and the eigenvalue equation for $g$. The resulting identity $(\lambda-\mu)(f,g)_{\mathrm{Pet}}=0$ forces orthogonality when $\lambda \ne \mu$.
[/proofplan]
[step:Use self-adjointness to show that the eigenvalues are real]
Because $f \ne 0$ and the Petersson inner product is positive definite on $S_k(SL_2(\mathbb{Z}))$, we have $(f,f)_{\mathrm{Pet}} > 0$. Using $T_n f = \lambda f$, self-adjointness of $T_n$, and conjugate-linearity in the second argument,
\begin{align*}
\lambda (f,f)_{\mathrm{Pet}}
&= (T_n f,f)_{\mathrm{Pet}} \\
&= (f,T_n f)_{\mathrm{Pet}} \\
&= (f,\lambda f)_{\mathrm{Pet}} \\
&= \overline{\lambda}(f,f)_{\mathrm{Pet}}.
\end{align*}
Since $(f,f)_{\mathrm{Pet}} > 0$, division gives $\lambda=\overline{\lambda}$, so $\lambda \in \mathbb{R}$.
The same computation with $g$ gives
\begin{align*}
\mu (g,g)_{\mathrm{Pet}}
&= (T_n g,g)_{\mathrm{Pet}} \\
&= (g,T_n g)_{\mathrm{Pet}} \\
&= (g,\mu g)_{\mathrm{Pet}} \\
&= \overline{\mu}(g,g)_{\mathrm{Pet}}.
\end{align*}
Because $g \ne 0$, $(g,g)_{\mathrm{Pet}} > 0$, hence $\mu=\overline{\mu}$ and $\mu \in \mathbb{R}$.
[guided]
We first record that the eigenvalues are real. This is not a separate hypothesis: it follows from self-adjointness.
Since $f$ is a nonzero cusp form and $(\cdot,\cdot)_{\mathrm{Pet}}$ is an inner product, its Petersson norm is positive:
\begin{align*}
(f,f)_{\mathrm{Pet}} > 0.
\end{align*}
Now compute the same scalar $(T_n f,f)_{\mathrm{Pet}}$ in two ways. The eigenvalue equation $T_n f=\lambda f$ gives
\begin{align*}
(T_n f,f)_{\mathrm{Pet}}=\lambda(f,f)_{\mathrm{Pet}},
\end{align*}
because the Petersson inner product is linear in the first argument. Self-adjointness gives
\begin{align*}
(T_n f,f)_{\mathrm{Pet}}=(f,T_n f)_{\mathrm{Pet}}.
\end{align*}
Substituting $T_n f=\lambda f$ into the second argument and using conjugate-linearity in that argument gives
\begin{align*}
(f,T_n f)_{\mathrm{Pet}}=(f,\lambda f)_{\mathrm{Pet}}=\overline{\lambda}(f,f)_{\mathrm{Pet}}.
\end{align*}
Combining these identities,
\begin{align*}
\lambda(f,f)_{\mathrm{Pet}}=\overline{\lambda}(f,f)_{\mathrm{Pet}}.
\end{align*}
The factor $(f,f)_{\mathrm{Pet}}$ is strictly positive, so it can be divided out. Hence $\lambda=\overline{\lambda}$, which means $\lambda \in \mathbb{R}$.
The same argument applies to $g$. Since $g \ne 0$, we have $(g,g)_{\mathrm{Pet}}>0$, and
\begin{align*}
\mu (g,g)_{\mathrm{Pet}}
&= (T_n g,g)_{\mathrm{Pet}} \\
&= (g,T_n g)_{\mathrm{Pet}} \\
&= (g,\mu g)_{\mathrm{Pet}} \\
&= \overline{\mu}(g,g)_{\mathrm{Pet}}.
\end{align*}
Dividing by $(g,g)_{\mathrm{Pet}}>0$ gives $\mu=\overline{\mu}$, so $\mu \in \mathbb{R}$.
[/guided]
[/step]
[step:Compare the two evaluations of $(T_n f,g)_{\mathrm{Pet}}$]
Using the eigenvalue equation $T_n f=\lambda f$ and linearity in the first argument,
\begin{align*}
(T_n f,g)_{\mathrm{Pet}}=\lambda(f,g)_{\mathrm{Pet}}.
\end{align*}
Using self-adjointness, the eigenvalue equation $T_n g=\mu g$, and conjugate-linearity in the second argument,
\begin{align*}
(T_n f,g)_{\mathrm{Pet}}
&= (f,T_n g)_{\mathrm{Pet}} \\
&= (f,\mu g)_{\mathrm{Pet}} \\
&= \overline{\mu}(f,g)_{\mathrm{Pet}}.
\end{align*}
Since $\mu \in \mathbb{R}$, $\overline{\mu}=\mu$. Therefore
\begin{align*}
\lambda(f,g)_{\mathrm{Pet}}=\mu(f,g)_{\mathrm{Pet}}.
\end{align*}
Subtracting the right-hand side from the left-hand side gives
\begin{align*}
(\lambda-\mu)(f,g)_{\mathrm{Pet}}=0.
\end{align*}
[guided]
We now compute $(T_n f,g)_{\mathrm{Pet}}$ in two different ways. The first computation uses the eigenvalue equation for $f$:
\begin{align*}
(T_n f,g)_{\mathrm{Pet}}=(\lambda f,g)_{\mathrm{Pet}}=\lambda(f,g)_{\mathrm{Pet}},
\end{align*}
where the scalar $\lambda$ leaves unchanged because the Petersson inner product is linear in the first argument.
The second computation moves $T_n$ from the first slot to the second slot. This is exactly the self-adjointness hypothesis:
\begin{align*}
(T_n f,g)_{\mathrm{Pet}}=(f,T_n g)_{\mathrm{Pet}}.
\end{align*}
Now use the eigenvalue equation $T_n g=\mu g$:
\begin{align*}
(f,T_n g)_{\mathrm{Pet}}=(f,\mu g)_{\mathrm{Pet}}.
\end{align*}
Because the Petersson inner product is conjugate-linear in the second argument,
\begin{align*}
(f,\mu g)_{\mathrm{Pet}}=\overline{\mu}(f,g)_{\mathrm{Pet}}.
\end{align*}
From the previous step, $\mu$ is real, so $\overline{\mu}=\mu$. Thus the second computation gives
\begin{align*}
(T_n f,g)_{\mathrm{Pet}}=\mu(f,g)_{\mathrm{Pet}}.
\end{align*}
Comparing the first and second computations,
\begin{align*}
\lambda(f,g)_{\mathrm{Pet}}=\mu(f,g)_{\mathrm{Pet}}.
\end{align*}
After subtracting the right-hand side, we obtain
\begin{align*}
(\lambda-\mu)(f,g)_{\mathrm{Pet}}=0.
\end{align*}
[/guided]
[/step]
[step:Use the distinctness of the eigenvalues to force orthogonality]
By hypothesis $\lambda \ne \mu$, so $\lambda-\mu \ne 0$. Since
\begin{align*}
(\lambda-\mu)(f,g)_{\mathrm{Pet}}=0,
\end{align*}
division by the nonzero scalar $\lambda-\mu$ gives
\begin{align*}
(f,g)_{\mathrm{Pet}}=0.
\end{align*}
This is exactly the asserted Petersson orthogonality of $f$ and $g$.
[/step]
