A semiclassical lecture note I am reading states that in the case of a planar domain (dimension $d = 2$), one has an $O(\sqrt{n})$ bound on the number of eigenvalues of the Laplacian operator $\Delta$ on the interval sequence of the form $[n - \epsilon, n + \epsilon]$ by using the Weyl's law $N(\lambda) = c_d\mathrm{vol}(M)\lambda^{d/2} + O(\lambda^{(d - 1)/2})$, where $N(\lambda)$ counts the number of eigenvalues of $\Delta$ less than or equal to $\lambda$, $M$ is the domain we are interested in (with some nice boundary conditions, e.g. Dirichlet) and $d$ is the dimension we are working in. With this asymptotic formula we have that
$$N(n + \epsilon) - N(n - \epsilon) = c_2\mathrm{vol}(M)(n + \epsilon) + O(\sqrt{n + \epsilon}) - c_2\mathrm{vol}(M)(n - \epsilon) - O(\sqrt{n - \epsilon}) = 2\epsilon c_2 \mathrm{vol}(M) + O(\sqrt{n + \epsilon}) - O(\sqrt{n - \epsilon})$$
I am not that familiar with Landau spaces (i.e. the big O), so how do we deduce from this the $O(\sqrt{n})$ bound for the number of eigenvalues?
What you have done looks good so far. Recall, we say that $f=O(g)$ if $\vert f(n) \vert \leqslant C \vert g(n) \vert$ for some $C>0$ provided that $n$ is sufficiently large.
Hence, if $f_1=O(\sqrt{n-\varepsilon})$ and $f_2 = O(\sqrt{ n + \varepsilon})$ then there exists $C_1,C_2>0$ such that $\vert f_1(n) \vert\leqslant C_1 \sqrt{n-\varepsilon}$ and $\vert f_2(n) \vert \leqslant C_2 \sqrt{n+\varepsilon}$ for $n$ large. Thus, \begin{align*}\vert 2 \varepsilon c_2 \mathrm{vol}(M) + O(\sqrt{n+\varepsilon}) - O(\sqrt{n-\varepsilon})\vert&\leqslant (2\varepsilon c_2 \mathrm{vol}(M) +C_1 \sqrt{n+\epsilon} + C_2 \sqrt{n-\varepsilon} \\ &\leqslant C \sqrt{n} \end{align*} for some constant $C>0$ for $n$ large. This tells us that $$N(n+\varepsilon)-N(n-\varepsilon) = O(\sqrt n) $$ as required.