Make a $52 \times 52$ grid. Let the columns be the first draw and the rows be the second draw. Label the columns, left-to-right, ace of spaces, 2 of spaces, 3 of spades, king of spades, ace of hearts, ... king of hearts, ace of diamonds, ..., king of diamonds, ace of clubs, king of clubs. Use the same labels in the same order for the rows.
We're going to compute the probability twice -- once wrongly and once rightly.
First wrongly. There are four columns corresponding to drawing a queen first. In those four columns, there are thirteen rows corresponding to drawing a diamond second. So there are $4 \cdot 13$ cells of our table that correspond to the desired sequence of cards. Happily $4 \cdot 13 = 52$ and there are $52 \cdot 52$ cells, so we have $1/52$ of the cells corresponding to satisfying draws.
We notice that the table is really four copies of the same four columnar blocks, where each block is one suit wide and runs from the top to the bottom of the table. If we were to stack these four copies onto one copy, the top row would just be the ranks (since all four suits of that rank landed in the same column when we stacked). We could also collapse the groups of rows having the same suit into single rows (because we do not care about the rank of the second card). Having done so, we have compressed the table to $13$ columns and $4$ rows and all the original cells corresponding to satisfying draws landed in one resulting cell, the one in the queen column and diamond row. That is, the compressed table can be used for counting to determine the odds of drawing the queen of diamonds in one draw.
Now the right way... But, we forgot a detail. A card that is taken in the first draw is not available in the second draw. So from our first table, we should delete the diagonal from the upper-left to the lower-right (from (ace of spades, ace of spades) to (king of clubs, king of clubs) ). Now every cell corresponds to different cards on the first and second draws. The only deleted cell corresponding to a satisfying draw is the (queen of diamonds, queen of diamonds) cell -- all other satisfying draws are still present. This means we have $4 \cdot 13 - 1$ surviving cells corresponding to satisfying draws and $52 \cdot 52 - 52$ surviving cells after all the deletions. The fraction of satisfying draws is $51/(52 \cdot 51) = 1/52$.
Why are these the same? Because the deletion process removed the same proportion of cells as it removed draw satisfying cells. But here's another way to see it. Suppose we marked the deleted cells, then stacked the same way we did in the first calculation. After stacking the columnar blocks, there are four diagonal stripes of marks corresponding to the deleted cells. Then, when we collapse the rows, each resulting cell has exactly one mark in it and each resulting cell receives one cell from each suit (columnar blocks) and one cell from each rank (row compression), so each of our $13 \cdot 4 = 52$ compressed cells corresponds to $52$ original cells and one of those original cells is marked. When we delete the marked (sub-)cell, we remove $1/52$ of the original cells landing in the draw satisfying cell and we remove $1/52$ of the original cells landing in every other cell -- we remove the same proportion of both, so their ratio is unchanged.
