Here's another way to think about tilings an Aztec diamond 
with skew tetrominoes (it's not really a new proof that no 
such tilings exist, but it's a new way to present the old 
one).  Suppose this region can be tiled by skew tetrominos.  
Take the 1-skeleton of the tiling, and color its edges with 
colors A, B, C, and D as shown below:

                o---B---o---A---o
                |       |       |
                D       C       D
                |       |       |
        o---B---o---A---o---B---o---A---o
        |       |       |       |       |
        D       C       D       C       D
        |       |       |       |       |
o---B---o---A---o---B---o---A---o---B---o---A---o
|       |       |       |       |       |       |
D       C       D       C       D       C       D
|       |       |       |       |       |       |
o---A---o---B---o---A---o---B---o---A---o---B---o
|       |       |       |       |       |       |
C       D       C       D       C       D       C
|       |       |       |       |       |       |
o---B---o---A---o---B---o---A---o---B---o---A---o
        |       |       |       |       |
        C       D       C       D       C
        |       |       |       |       |
        o---B---o---A---o---B---o---A---o
                |       |       |
                C       D       C
                |       |       |
                o---B---o---A---o

Now shrink all the segments colored A and D down to length 0.
You can check that every tile-boundary is a closed (possibly 
degenerate) polygon that, if superimposed on a checkerboard, 
encloses equal numbers of black and white squares.  But the 
boundary of the Aztec diamond shrinks down to

o---B---o---B---o---B---o
|                       |
C                       C
|                       |
o---B---o               o
        |               |
        C               C
        |               |
        o-------o       o
                |       |
                C       C
                |       |
                o---B---o

which does not enclose equal numbers of black and white
squares.

This proof sweeps a few details under the rug (such as
the way the possibly degenerate polygons fit together),
and one still is using a hidden argument about winding
numbers.  But it's still essentially a complete proof.