I need to prove, by induction, that for all $n$ there exists an $m$ with the property that $$m^2 \leq n \leq (m+1)^2$$
I can easily establish a base case (picking $n = m = 0$). I am finding it harder to assume this property holds and find an $m$ that makes it true for $n+1$.
Any hints very appreciated.
For the induction step using your approach, suppose the result is true for $n$. Show it is true for $n+1$.
Let $a$ be a number that "works" for $n$. Maybe we get lucky and $n+1\le (a+1)^2$, and we are finished. We can take $m=a$.
If we are unlucky and $(a+1)^2\lt n+1$, then note that $n+1\le (a+2)^2$. Thus in this case we get $(a+1)^2\le n+1\le (a+2)^2$, and we can take $m=a+1$.