Prove that the equation $x^2=y$ has a unique solution

What you have shown is that given a real number $y > 0$ there is at most one $x > 0$ such that $x^2 = y$. You are yet to show that such an $x$ exists for every $y > 0$. For this you will need to use the completeness of the reals somewhere, as the answers to the linked question indicate. For instance, your proof works just as well over the rationals: as per your proof, for every rational $y > 0$ there is at most one rational $x > 0$ such that $x^2 = y$. But it is also possible that there is no such rational $x$ for some values of $y$ (for instance, take $y = 2$).

So, your strategy covers the uniqueness of the square root, whereas the existence of the square root still remains to be proved.