Because it’s one queen per column and one queen per row, these regions necessarily have a queen on the intersection of the only row and only column possible.
Initial board example:
After eliminating squares:
Whether it’s a one-square-large region, or a region that has had all other squares Xed-out, there’s only one place to put a queen.
Likewise with rows and columns.
Each row or column needs one queen, so if a region spans all possible squares of a row or column, a queen must be there and not in the rest of the region.
Example: These two squares would block all of the green region’s squares due to covering three by proximity, and the fourth by row.
Example: These two regions have their only possible squares on two rows. Placing a queen on any other square of these two rows would make it impossible to have enough rows for these two regions. The same applies when there are three, four or more regions that have their only possible squares on three, four or more rows or columns. Bear in mind that the rows or columns may not be next to each other.
Example: These three columns must contain three queens in the three regions available. All other squares in these three regions can be safely eliminated.
The process of placing Xs will reduce possible squares in regions, and fewer possible squares in regions will reveal new squares to place an X or queen.
If it’s a dead end, better to hit it early.
Process of elimination.
P.S. I’m aware that in my examples, I made some wrong placements. They’re examples.