Two equal critical levels can be separated by adding disjoint bump perturbations near the corresponding critical points

The partial derivatives are $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x=-2\pi \sin (2\pi x)$ and $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y=-2\pi \sin (2\pi y)$, so the critical points are exactly the four points with $x,y\in \{0,\frac{1}{2}\}$ modulo $\mathbb{Z}$. At the two saddles $(0,\frac{1}{2})$ and $(\frac{1}{2},0)$ the critical value is $0$.

The Hessian is diagonal with entries $-4{\pi }^2\cos (2\pi x)$ and $-4{\pi }^2\cos (2\pi y)$, so all four critical points are nondegenerate. By [F1], $f$ is Morse. Choose pairwise disjoint neighbourhoods of the two saddles and bump functions ${\rho }_1,{\rho }_2$ that are identically $1$ near the corresponding saddle and supported away from the other one.

For sufficiently small $\epsilon >0$, define $g:=f+\epsilon {\rho }_1-\epsilon {\rho }_2\mathrm{.}$ The compact perturbation argument from [L1] applies to these two fixed bumps: for small enough $\epsilon$, the function $g$ has the same critical points as $f$ and the same Hessians at those critical points. Near the first saddle one has $g=f+\epsilon$, and near the second one has $g=f-\epsilon$, so the two saddle critical values become $\epsilon$ and $-\epsilon$.

Hence two equal critical levels can be separated by disjoint bump perturbations without changing the local Hessians.