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.