If $V$ is a real vector space with an inner product, we can use the Gram-Schmidt process to obtain an othonormal basis. Under this basis, the matrix $A$ associated to the linear operator $T$ will be a matrix with real entries such that $A=A^{\mathrm{T}}$. The point is that to study $T$, we only need to study $A$. So, we have transformed our question into a question regarding matrices. But from the answer to the question you cited, viewing $A$ as a complex matrix (then $A$ as a complex matrix is normal having only real eigenvalues), we conclude that $A^{*}=A$. This means that $A$ is Hermitian ($A^\mathrm{T}=A$) and thus $T$ is Hermitian.
In other words, your question is really only a question about matrices. Although if you want a more abstract point of view, what we are doing here is in fact working with the complexification of the inner product space to obtain a result about the original operator on the real inner product space.
Addendum:
If you really feel uneasy about treating real matrices as complex matrices, which you shouldn't feel uneasy about, and insist that everything be done in the real domain. Here is another way to prove it without using complex numbers.
The basic strategy is to use Schur's Decomposition Theorem for real matrices with real eigenvalues.
Suppose that $A$ is a real matrix with all of its eigenvalues being real numbers, then there is an orthogonal matrix $Q$, and an upper triangular real matrix $U$ such that $A=QUQ^{\mathrm{T}}$.
Proof. By induction on the size of $A$ (Cf. here). Let $\lambda_1$ be an eigenvalue of $A$, and $\pmb{x}_1\in \mathbf{R}^n$ an eigenvector corresponding to $\lambda_1$. Extend $\pmb{x}_1$ to an orthonormal basis $\pmb{x}_1, \pmb{x}_2, \ldots, \pmb{x}_n$. Let $Q_1$ be the matrix whose $i$th column is the vector $\pmb{x}_i$. Then we have
$$ AQ_1=A(\pmb{x}_1, \pmb{x}_2, \ldots, \pmb{x}_n) = (\pmb{x}_1, \pmb{x}_2,\ldots,
\pmb{x}_n)
\begin{pmatrix}
\lambda_1 & \pmb{*} \\
\pmb{0} & A_1
\end{pmatrix}
= Q_1
\begin{pmatrix}
\lambda_1 & \pmb{*} \\
\pmb{0} & A_1
\end{pmatrix}
.
$$
Since $A_1$ has a smaller size, by induction, there exists an $(n-1)\times (n-1)$ orthogonal matrix $Q_2$ such that $A=Q_2U_2Q_2^\mathrm{T}$ where $U_2$ is an upper triangular matrix. Therefore we have
$$ AQ_1 =Q_1
\begin{pmatrix}
\lambda_1 & \pmb{0} \\
\pmb{0} & Q_2U_2Q_2^\mathrm{T}
\end{pmatrix}=Q_1
\begin{pmatrix}
1 & \pmb{0} \\
\pmb{0} & Q_2
\end{pmatrix}
\begin{pmatrix}
\lambda_1 & \pmb{0} \\
\pmb{0} & U_2
\end{pmatrix}
\begin{pmatrix}
1 & \pmb{0} \\
\pmb{0} & Q_2
\end{pmatrix}^\mathrm{T}.$$
Now let
$$Q=Q_1
\begin{pmatrix}
1 & \pmb{0} \\
\pmb{0} & Q_2
\end{pmatrix}\quad
\text{and}\quad
U=
\begin{pmatrix}
\lambda_1 & \pmb{0} \\
\pmb{0} & U_2
\end{pmatrix}.
$$
Then $U$ is upper triangular and $Q$ is orthogonal and $A=QUQ^\mathrm{T}$.
After you prove the decomposition theorem, we can then show that if $A$ corresponds to your linear operator $T$, then $A=QUQ^{\mathrm{T}}$ and so using $AA^\mathrm{T}=A^\mathrm{T}A$, we get that $UU^{\mathrm{T}}=U^\mathrm{T}U$.
Suppose that $U$ is an upper triangle matrix with $UU^\mathrm{T}=U^\mathrm{T}U$, then $U$ is diagonal.
Proof. Denote the $ij$-entry of $U$ by $u_{ij}$. Then the for any $i$, we have
$$ \sum_{j=1}^n u_{ij}u_{ij} = \sum_{j=1} u_{ji}u_{ji}\implies \sum_{j=1}u_{ij}^2 =\sum_{j=1}^n u_{ji}^2. \tag1$$
But because $U$ is upper triangular, Let $i$ be $1$ in Eq.(1), then we get that on the first row of $U$, every entry except the one on the first column is zero. Now, let $i$ be $2$ in Eq. (1), we immediately get that on the second row of $U$, every entry except the one on the second column is zero. This carries on. So $U$ has to be diagonal.
Now, since $U$ is diagonal, it follows that $A^\mathrm{T}=QU^\mathrm{T}Q^\mathrm{T}= QUQ^\mathrm{T}=A$. This shows that $A$ is indeed Hermitian.
