This question already has an answer here:
hint: $\dfrac{1}{\sqrt{k}}<2\sqrt{k}-2\sqrt{k-1}$. You let $k$ runs from $1$ to $n$, and add the inequalities up.
If the proposition holds true for $n=m$
$$\sum_{r=1}^m\dfrac1{\sqrt r}\le2\sqrt m-1$$
$$\implies\sum_{r=1}^{m+1}\dfrac1{\sqrt r}=\dfrac1{\sqrt{m+1}}+\sum_{r=1}^m\dfrac1{\sqrt r}\le\dfrac1{\sqrt{m+1}}+2\sqrt m-1$$
It if sufficient to show $$\dfrac1{\sqrt{m+1}}+2\sqrt m-1\le2\sqrt{m+1}-1$$
$$\iff\dfrac1{\sqrt{m+1}}\le2[\sqrt{m+1}-\sqrt m]=\dfrac2{\sqrt{m+1}+\sqrt m}$$
$$\iff \sqrt{m+1}\ge\sqrt m$$ which is true
Without induction you could use this inequality
$$\frac1{\sqrt k}\le2(\sqrt{k}-\sqrt{k-1})$$
and then you telescope. Notice also that the same inequality is useful for your attempt.