Why can we drop the superscripts in the Floyd Warshall algorithm?
Basically in iteration k no entry $d_{ij}$ can change if $i=k$ or $j=k$. Since you update the entries in $D$ by considering $d_{ik}$ and $d_{kj}$ you can do all these updates in-place.

However there is an even simpler argument if you only want to show that the algorithm requires $O(n^2)$ space. At every iteration $k$ the values in $D^k$ only depend on the previous matrix $D^{k-1}$. Therefore it is enough to store the matrix $D^{k-1}$ of the previous iteration, use it to fill the next matrix $D^k$ and then you can safely discard $D^{k-1}$. So you only ever need to keep two matrices in memory and arrive at $O(n^2)$ space. Of course no sensible implementation would actually do this but it is enough to show the space bounds.