You actually get |(x-1)|>=0.

Meaning x-1 >= 0 or 1-x >=0 so x >=1 or  1 >= x.

The square root of x^2 is |x|, not just x
(-4)^2 > 0

Take the square root of both sides:

-4 > 0

You have to be more careful than that.
not every operation that you can do to an equation will also work if you do it to an inequality, for example taking the square root. functions that preserve inequalities (i.e. functions f where f(x) > f(y) implies x > y) are called increasing functions, and there is no reason to believe that an arbitrary function that you encounter will be increasing. some functions are, like f(x)=x^(3) and f(x)=log(x), but most are not, like f(x)=x^(2) (unless you restrict the domain to the positive reals, in which case it is).
In addition to what others have pointed out, an equation like $f(x)$\^2 >= 0 is true for all x, since the square of any real number is non-negative.

So as soon as you get to the form (x - 1)\^2 >= 0, you know the solution set is all real numbers.
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