Suppose $0
Let a and b be real numbers, and suppose $0<a<b$. 
Prove for all $n\geq 2$,  $0< \sqrt[n]a< \sqrt[n]b$.
Proof: Suppose there exists an $n\geq 2$ such that $0 \geq \sqrt[n]a \text{ or } \sqrt[n]a \geq\sqrt[n]b$. 
case $0 \geq \sqrt[n]a $: 
Then $0 \geq a$, which contradicts our assumption that $0<a<b$. 
case $\sqrt[n]a \geq\sqrt[n]b$ Then
$a \geq b$, which contradicts our assumption that  $0<a<b$. 
In all cases, there is a contradiction with the assumption that $0<a<b$. Therefore, the proposition holds.

Is the above proof correct? Can this statement be proven by induction on the natural number?

You can prove this by showing that $\sqrt[n]{x}$ is monotonically increasing when $n\geq 2$. 
Let $f(x)=\sqrt[n]{x}$, then $f'(x)=\frac{1}{n}x^{\frac{1-n}{n}}>0$ when $n\geq 2$, so $f(x)$ is monotonically increasing when $n\geq 2$. Also, $f(x)>0$ when $x>0$. Therefore, for all $n \geq 2, 0<\sqrt[n]{a}<\sqrt[n]{b}$.
