How can the moves from this pawn endgame study be found without a computer? [FEN "8/1p5k/1P1p4/3p4/3Pp2p/2K1P2p/7P/8 w - - 0 0"]
[PlyCount "21"]

1. Kb2 Kg8 2. Ka1 Kf8 3. Ka2 Ke7 4. Kb3 Kd7 5. Kb4 Kc6 6. Ka5
Kd7 7. Kb4 Ke6 8. Kc3 Kf6 9. Kc2 Kg5 10. Kd1 Kg4 11. Ke2 


In this great study from Ebersz, white can reach a draw, but only with
extreme accurate play. The given variation is one possible. All white 
moves are forced, any other king move would lose. Suppose, white is
in such a position in a practical game, of course without access to
electronic topics. Can the correct squares be found out manually ?
I know the theory of countersquares, but it seems to be extremely
difficult.

How can white save this game without too much efforts ?
Some of the corresponding squares are quite easy to work out, especially if you write things down (not allowed in a real game).

First of all, it is clear that

a5 corresponds to c6.

Black can move from c6 to g4 in 4 moves. The only square that stops black's invasion and is 4 moves away from a5 is e2, so 

e2 corresponds to g4.

The squares on the paths in between need to correspond, so

b4 corresponds to d7,

c3 corresponds to e6,

d2 corresponds to f5.

It then follows that

b3 corresponds to e7,

c2 corresponds to f6,

d1 corresponds to g5.

Hence,

b2 corresponds to f7,

c1 corresponds to g6,

b1 corresponds to g7.

The trickiest part is to show that a3 corresponds to e8. When the black king is on e8, it threatens to go to d7, e7 or f7, corresponding to b4, b3 and b2 for white. Therefore the white king needs to be on a3 or c3, but c3 is the wrong square. Why? Because after a subsequent Kd8 move form black, white would be in zugzwang. Black can still go to d7or e7, so white needs to go to a square adjacent to b4 and b3. Such a square is not available from c3.

So,

a3 corresponds to e8,

and therefore

a4 corresponds to d8,

a2 corresponds to f8

a1 corresponds to g8.

Now the solution can be explained. In the diagram position Black can go to g6, g7 or g8, so white needs to have c1, b1 and a1 available. Therefore 1. Kb2!