Flawed proof of theorem 161 in E. Landau's Foundations of Analysis? 
Theorem 161:  For each ζ, the equation ξξ = ζ has exactly one solution. 

In part II of the proof the author considers the set of all rational numbers for which X~X~ &LT; ζ. And claims that it constitutes a cut. Where X is a rational number (class of equivalent fractions) . And X~ is the set of all rational numbers &LT; X. Which is a cut by theorem 150.  Rational cuts have been renamed rational numbers at this point in the text due to the results of chapter III section 5. 
To prove property 2 of cuts it suffices to show:

With every number it contains, the set also contains all numbers smaller than that number.

The author proves property 2 in the following way:

If X~X~ &LT; ζ, Y~ &LT; X~ then Y~Y~ &LT; X~X~ &LT; ζ

However if we consider a ζ of 10~ say. An X~ of 3~ say. And a Z~ of 2~ &LT; X~X~. Then by theorem 162 (√2 is irrational) our Z~ cannot be of the form R~R~ = Z~. Therefore X~X~ &LT; ζ does not constitute a cut, by counterexample.

Your confusion：

"Because I can come up with a number less than a number belonging to the set which  cannot be expressed as the square of a rational number. So it does not belong to the set. Property 2 of cuts is not satisfied. "

My answer:
① property 2 said
    "With every number it contains, the set also contains all numbers smaller than that number."
   By definition 28: A set of rational numbers is called a cut if ......
That means: until now, any numbers we talk about must only be rational numbers.
② so，even $\sqrt{2}$ which is less than a number belonging to the set（but $\sqrt{2}$ itself not belong to the set ）is still ok,
   because $\sqrt2$ satisfied （less than a number belonging to the set)，but not satisfied （itself need to be rational).
③ until definition 42，the book do not think $\sqrt{2}$ as a number，but as a cut.
