About compact region that is homeomorphic to a disk in a torus. Let $C_1$ be a circle embedded in a standard torus $T$. Let $p$ and $q$ be two points of $C_1$. Assume that $C_2$ is a circle in $T$ distinct from $C_1$ such that $C_1$ and $C_2$ intersect at $p$ and $q$. We orient the two circles. In $C_1$, denote the arc from $p$ to $q$ by $l_1$ and from $q$ to $p$ by $l_2$. In $C_2$, denote the arc from $p$ to $q$ by $l_3$ and from $q$ to $p$ by $l_4$. We obtain three compact regions (1) region bounded by $l_1 \cup l_3$, (2) region bounded by $l_3 \cup l_2$ and (3) region bounded by $l_2 \cup l_4$. Suppose that one of the three regions above is with a hole ,i.e it is not a disk indeed. Then is it possible that one of the two remaining regions (or both of them) is (are) also not homeomorphic to a disk.   

Suppose that $C_2$ encircles a side on the torus, while $C_1$ is a simple circle that happens to intersect it. Then regions (1) and (2) are simple collapsible subsets of the interior of $C_1$, while region (3) is the entire torus and so is not a disk.
Switch them around: $C_1$ encircles a side, while $C_2$ is simple. Regions (1) and (3) are still the same, but region (2) is also the entire torus.
For a third example, suppose both $C_1$ and $C_2$ both encircle a side, but one of them is angled so that they cross at the top and bottom. If we choose $l_1, l_3$ are the segments on the inside of the torus and $l_2, l_4$ are on the outside, then (1) and (3) are disks, while (2) is the entire torus. If instead the circles are oriented to make $l_1, l_4$ on the inside of the torus and $l_2, l_3$ on the outside, then regions (1) and (3) are the entire torus and region (2) is a disk.
I've yet to spot a scenario where all three regions are the entire torus, but I would not be surprised if one is possible.
