We read the assertion for left to right. You are trying to show that there exists a number $a$ which has a certain property. What property? That whatever number $b$ is, if $a\lt b$ then $a^2\lt b^2$.
If you can show that (say) $10$ has the property, you will be completely finished.
To show that there exists an $a$ with a certain property, all you need to do is toexhibit an $a$ that has the property. (There may also be indirect ways to show such an $a$ exists.)
So is it true that for any (all) $b$, if $10<b$, then $100<b^2$? Sure, on the positive reals, $f(x)=x^2$ is an increasing function.
Remark: Pick $a=-17$. Then $a$ does not have the required property. For $-17 \lt 2$, but $(-17)^2 \gt 2^2$. So there are "bad" $a$. But the statement just claimed that there is a good $a$.
You might note that we could have picked $a=0$, for $0$ has the desired property. Actually, nothing smaller than $0$ has the property.