$0I want to show that given that $x,y>0$, we can deduce that $0<x^2<y^2 \Rightarrow x<y$. I am having problems with squarerooting the inequalities here.
$\sqrt{x^2}=|x|$ and $|x|=x$ here since x is positive so can we just squareroot the double inequality and deduce that the implication holds here.
In other words $$0<x^2<y^2 \Rightarrow 0<|x|<|y| \Rightarrow 0<x<y$$ and hence we get our result. Is this valid?

Note that you actually translate the problem of "$(\cdot)^2$ being increasing" to "$\sqrt{\cdot}$ being increasing" or vice verca, depending on how you look at it. You say

we just squareroot the double inequality and deduce

Yes, this only holds if the squareroot is increasing as well.
These two are equivalent. Indeed, let $f:A\to B$ for some subsets $A,B\subseteq\mathbb{R}$. 
Lemma 1. Assume that $f$ is invertible. Then $f$ is increasing if and only if $f^{-1}$ is increasing.
Proof. Indeed, let $x<y$. Since $f$ is invertible then $x=f(x'), y=f(y')$ for some $x',y'\in\mathbb{R}$. Note that $x'<y'$ because if $y'<x'$ then $f(y')<f(x')$ since $f$ is increasing. Contradiction because $f(y')=y$ and $f(x')=x$.
Now since $x'=f^{-1}(x)$ and $y'=f^{-1}(y)$ then we obtain that $x<y$ implies $f^{-1}(x)<f^{-1}(y)$.
So we've proved that $f$ is increasing $\Rightarrow$ $f^{-1}$ is increasing. The other implication comes simply by substituting $f$ with $f^{-1}$. $\Box$

So your approach is not really good. The implication
$$0<x^2<y^2 \Rightarrow x<y$$
actually means $\sqrt{x}$ is increasing. But you use that property later in the proof!!! This is wrong!
So as a first step I'm going to show that $(\cdot)^2$ is increasing without using circular references.
Let $0<a$. If $b<c$ then by general property of reals $ab<ac$.
So let $0<x<y$. Since $0<x$ then $xx<xy$. Since $0<y$ then $xy<yy$. Combining these two inequalities we obtain $xx<yy$, i.e. $x^2<y^2$. This implies that $(\cdot)^2$ is increasing on positive reals.
So now $\sqrt{\cdot}$ is increasing by Lemma 1 and the original implication follows.
