Can an Perron eigenvector of a non-symmetric irreducible nonnegative matrix be also a Perron eigenvector of its transpose?

Your contradiction approach will fail because the claim is not false. In terms of simple examples: consider any doubly stochastic matrix, e.g. a permutation matrix and in particular one that is not symmetric (I’d suggest the cyclic shift matrix).

Another argument for why the stated claim is false.

The space of $n \times n$ irreducible Perron matrices $X$ has dimension $n^2$ (if you want, restrict yourself to positive matrices).

Fix $\rho >0$ and $p$ a positive vector. The space of $n \times n$ irreducible Perron matrices $X$ such that $Xp = \rho p$ has dimension $n^2-n$ (we have n linear equations to check).

The space of $n \times n$ irreducible Perron matrices $X$ such that $Xp = \rho p$ and $p^t X = \rho p^t$ has dimension $n^2-2n+1 = (n-1)^2$ (we have $n$ more linear equations to check, the last one being redundant).

The space $n \times n$ irreducible symmetric Perron matrices $X$ such that $Xp = \rho p$ is included in the later, and has dimension $n(n+1)/2-n = n(n-1)/2$. If $n>2$, then $n(n-1)/2 < (n-1)^2$, so that there are non-symmetric Perron matrices with main eigenvector $p$ and main eigen-covector $p^t$.