We can use Taylor series to expand $\sqrt{x}$ $$ \sqrt{x} = \sqrt{n} + \frac{1}{2\sqrt{n}}(x-n) - \frac{1}{4n^{3/2}}(x-n)^2 + ... $$
The solution provided to me uses expansion point around $\sqrt{n}$. And then we are supposed to substitute $x=n+1$
$$ \sqrt{n+1} = \sqrt n + \frac{1}{2\sqrt n}(n + 1 - n) + ... $$
And we arrive at
$$ \sqrt{n+1} - \sqrt n = (\sqrt n + \frac{1}{2\sqrt n}(n + 1 - n) + ...) - \sqrt n \\ = O(\frac{1}{\sqrt n}) $$
What I don't understand is how the expansion point is chosen.
You can expand the series at anywhere you like (that make sense). For example, you can even try at $\sqrt{n + 1}$:
$$\sqrt{x} = \sqrt{n+1} + \dfrac{1}{2\sqrt{n+1}} (x - n - 1) + \dfrac{1}{4(n+1)^{3/2}}(x - n - 1)^2 + \cdots$$
and sub $x = n$:
$$\sqrt{n} = \sqrt{n+1} + \dfrac{1}{2\sqrt{n+1}} (-1) + \cdots$$
$$\Rightarrow \sqrt{n+1} - \sqrt{n} = \dfrac{1}{2\sqrt{n+1}} + \cdots = O(1/\sqrt{n+1}) \rightarrow O(1/\sqrt{n})$$
That $n$ goes to infinity as you stated.
It's just for your convenience to expand the series at $x = n$.