Initial comment: Begin by noting that, for all $n\geq 1$, we have that
$$
n(\sqrt{n}-2)+2>0\Longleftrightarrow n\sqrt{n}-2n+2>0\Longleftrightarrow \color{red}{\sqrt{n}>2-\frac{2}{n}}.\tag{1}
$$
Thus, it suffices for us to prove the proposition $P(n)$ for all $n\geq 1$ where
$$
P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\geq \sqrt{n}.\tag{2}
$$
If we can prove $(2)$, then we will have proven
$$
\color{blue}{\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}}\color{red}{\geq\sqrt{n}}\color{blue}{> 2-\frac{2}{n}},
$$
as desired. I'm sure you can handle the proof of $(1)$ quite easily.
Claim: For $n\geq 1$, let $P(n)$ denote the statement
$$
P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\geq \sqrt{n}.
$$
Base step: $P(1)$ holds since $1\geq\sqrt{1}$ is true.
Before induction step: Consider the following inequality for any $x\geq 1$:
$$
\sqrt{x}+\frac{1}{\sqrt{x+1}}>\sqrt{x+1}\tag{3}.
$$
Briefly, observe that for $x\geq 1, \sqrt{x(x+1)}>x$; thus, $\sqrt{x(x+1)}+1>x+1$. Dividing by $\sqrt{x+1}$ proves $(3)$. The purpose of $(3)$ is to streamline the calculations below in the inductive step.
Inductive step: Fix some $k\geq 1$ and suppose that $P(k)$ is true. Then
\begin{align}
\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} &\geq \sqrt{k}+\frac{1}{\sqrt{k+1}}\tag{by $P(k)$}\\[1em]
&> \sqrt{k+1},\tag{by $(3)$}
\end{align}
which shows that $S(k+1)$ follows. This concludes the inductive step.
Thus, for all $n\geq 1, P(n)$ is true. $\blacksquare$
