For any $r \in [1, n]$, we have $ \frac{1}{\sqrt{r}} \geqslant \frac{1}{\sqrt{n}}. $ with strict inequality for $1 \leqslant r \lt n$. Adding all these $n$ inequalities, $ \sum_{r=1}^{n} \frac{1}{\sqrt{r}} \gt n \cdot \frac{1}{\sqrt{n}} = \sqrt{n}. $
Proof using induction. The OP requested a proof using induction. I am assuming you can handle the base case $n=2$.
Now, assume that the inequality holds for some $n \geqslant 2$; we will verify the inequality for $n+1$: $ \begin{align*} \sum_{r=1}^{n+1} \frac{1}{\sqrt{r}} &= \sum_{r=1}^{n} \frac{1}{\sqrt{r}} + \frac{1}{\sqrt{n+1}} \\ &\gt \sqrt{n} + \frac{1}{\sqrt{n+1}} \\ &= \sqrt{n+1} + \frac{1}{\sqrt{n+1}} - (\sqrt{n+1} - \sqrt{n}) \\ &= \sqrt{n+1} + \frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &\gt \sqrt{n+1} , \end{align*} $ which is what we want to show.
Notice that out second inequality is a bit too crude. robjohn's answer shows how to get a better bound by being more careful.