An elementary combinatorial question inspired by G-C Rota's basis conjecture Let $X$ be a set, and let there be given $n$ different subsets $S_1$,...,$S_n$ of $X$ (different means no two of them are equal as sets, though their intersection is allowed to be non-empty), with each $S_i$ of cardinality $n$. Can one always totally order each of the sets $S_i$, in such a way that each $k$-th transverse set $T_k$ has cardinality $n$ (i.e. contains exactly $n$ distinct elements of $X$) for all $1 \leq k \leq n$, where we define $T_k$ as the subset of $X$ whose elements consist of the $k$'th smallest element of each of the now ordered sets $S_i$, as $i$ runs from $1$ to $n$?
Edit: inspired by amrsa's comment that the conclusion holds even if all sets $S_i$ are equal, it is tempting to ask the same question for any $n$ subsets $S_i$ of $X$ such that $|S_i|=n$ for all $1 \leq i \leq n$ (allowing for some of the $S_i$s to be equal as sets).
 A: Yes. We order the sets by a two-stage process, as follows. In the first stage, step $0$ is to array the $n$ sets as the $n$ rows of an $n\times n$ matrix, in arbitrary order. At step $k\;$ ($k=1,...,n-1$) of the first stage, in each row $l\;$ ($l=k+1,...,n$) of the matrix, any element of the row that is the same as an element—say the $j$th element—of the $k$th row swaps its position with the element in the $j$th position of the $l$th row, so that now the same element appears in the $j$th position of both the $l$th and the $k$th row. After the $n$ steps of the first stage are completed, in the second stage, the elements of each row $k\;$ ($k=1,...n$) are cycled forward by $k-1$ positions: thus the element in the $i$th position ($i=1,...,n$) is moved to position $i+k-1$ (modulo $n$).
It can be seen that, following this process, the $n$ elements in each column are distinct, whatever commonality there is between the given sets. 
The following answer addressed the question put in its original form, which was not the question intended by the OP.
Pick any element from the first set, and delete it from the remaining sets, leaving at least $n-1$ elements in each. Do the same again with the $n-1$ reduced sets, and so on. Now we have $n$ elements, one from each set. Pick any ordering of the elements in the sets such that the first element in the first set is the element we chose, ... , and the last element of the last set is the element we chose from that set.
In this construction, note that the second element we chose differs from the first one, because it was picked from a set that did not contain the first one; similarly, the next element differs from the first two, because it was picked from a set that did not contain those two; and so on. A stage $k$ of the process, an element is picked from a set containing at least $n-k$ elements that does not contain any of the previous $k$ selected elements. (Stage $0$ is the arbitrary selection of an element of the first set.) Thus, at stage $n-1$, a total of $n$ elements have been chosen, each differing from all the previously chosen elements. That is, the final set of selected elements has $n$ elements.
