Consider some column $i$ and some rows $j < k$. Let $x$ be the element at row $j$ column $i$ after the final sorting, and let $y$ be the element at row $k$ column $i$. We want to prove that $x \leq y$.
Since $x$ ended up in column $i$, there must have been $n-i$ elements in row $j$ larger than $x$. Together with $x$ itself, these are $n-i+1$ elements in row $j$ that were at least as large as $x$. The corresponding elements in row $k$ were also at least as large as $x$, since the columns were sorted. Summarizing, row $k$ contains at least $n-i+1$ elements which are at least as large as $x$. That means that after sorting, the element at position $i$ on row $k$ (that is, $y$) is at least as large as $x$