The problem is a bit confusing in why it choose to use skew-symmetric matrices. The claim in-fact holds for all non-symmetric matrices. I will write out the general proof while answering the questions you had.
Proposition: Let $A$ be a symmetric matrix. Then $A$ has a repeated eigenvalue if and only if there exists a non-symmetric matrix $J$ which commutes with $A$.
Proof: We begin with a generic matrix $M$. $M$ is any matrix, it doesn't matter which one.
A given diagonal matrix $D$ will commute with $M$ if and only if $d_im_{ij} = d_jm_{ij}$ This is a simple consequence of matrix multiplication as explained in your proof. Remember that the entries of the matrix are just numbers. Ordinary real numbers commute, so you have $d_jm_{ij} = m_{ij}d_j$, there is no problem there.
Continuing, if we have $d_im_{ij} = d_jm_{ij}$ then we must have $m_{ij}(d_i - d_j) = 0$. One of the two factors must be zero for this to happen, so either $m_{ij} = 0$ or $d_i = d_j$.
The third sentence in your proof is wrong. What we have established so far is that if $m_{ij} \neq 0$, then $M$ can commute with $D$ only if $d_i = d_j$. This is not and if an only if.
If $M$ commutes with $D$ and if $M$ has non-zero off diagonal entries then we must have repeated diagonal entries on $D$. In general, the more non-zero off diagonal entries that $M$ has, the more diagonal entries needs to be repeated for $D$. If $m_{ij}$ is non-zero for all off diagonal entries then we must have $d_i = d_j$ for all $i\neq j$ where then the matrix $D$ is just a scalar multiple of the identity. Clearly then having some random subset of diagonal entries repeated is not enough to guarentee that $D$ and $M$ commute; it needs to be a rather specific.
What is true however, is that there will always exist a $M$ with non-zero off-diagonal entries which commutes with $D$ when $D$ has repeated entries. If $d_i = d_j$ then simply take $m_{ij}$ and $m_{ji}$ to be whatever non-zero numbers you want and make the rest of the off-diagonal entries zero.
So far our claim is: There exists a matrix $M$ with non-zero off-diagonal entries that commute with $D$ if and only if $D$ has repeated entries.
To continue on with our proof, take some symmetric matrix $A$. Orthogonally diagonalize $A$ as $A = QDQ^\mathrm{T}$ Suppose then that some non-symmetric matrix $J$ commutes with $A$. Then take $M = Q^\mathrm{T}JQ$ We know that $M$ is asymmetric which necessarily means that there exists non-zero off-diagonal entries. $M$ and $D$ commute, so $D$ must have repeated eigenvalues.
Conversely, suppose that $D$ has repeated eigenvalues. From our previous discussion, we can construct a matrix $M$ which commutes with $D$. More precisely, we are free to choose $M$ to be non-symmetric. Then $J=QMQ^\mathrm{T}$ is a non-symmetric matrix which commutes with $A$.