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Edit2: everything works, updating the answer.

Yes. Consider two cases.

Case 1. There is a point on the boundary contained both in the closure of $U$ and $V$. Do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.

Now take the map $(x, y) \mapsto x+y$.

It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.

Case 2. One of the subsets (say $U$ without loss of generality) contains $\partial\mathbb{D}$ in it's closure. Then, it contains a small annulus $1-\varepsilon<|r|<1$. There is a map from disk to itself which is surjective from this annulus: one can do such an automorphism of a disk that $0$ is not in the annulus anymore and then use $z \mapsto z^2$. Composing this map with the projection on the first component we obtain the desired result.

Edit2: everything works, updating the answer.

Yes. Consider two cases.

Case 1. There is a point on the boundary contained both in the closure of $U$ and $V$. Do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.

Now take the map $(x, y) \mapsto x+y$.

It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.

Case 2. One of the subsets (say $U$ without loss of generality) contains $\partial\mathbb{D}$ in it's closure. Then, it contains a small annulus $1-\varepsilon<|r|<1$. There is a map from disk to itself which is surjective from this annulus: one can do such an automorphism of a disk that $0$ is in the image of the annulus and then use $z \mapsto z^2$. Composing this map with the projection on the first component we obtain the desired result.

Edit: actually, proof works in the assumption that $U$ and $V$ contain a point on the boundary (for example if they are simply-connected and non of them is a whole $\mathbb{D}$). I do not know whether the answer is "yes" for the covering by an annulus and a concentric circle.

(Yes). Consider a point on the boundary contained both in the closure of $U$ and $V$ and do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.

Now take the map $(x, y) \mapsto x+y$.

It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.

Edit2: everything works, updating the answer.

Yes. Consider two cases.

Case 1. There is a point on the boundary contained both in the closure of $U$ and $V$. Do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.

Now take the map $(x, y) \mapsto x+y$.

It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.

Case 2. One of the subsets (say $U$ without loss of generality) contains $\partial\mathbb{D}$ in it's closure. Then, it contains a small annulus $1-\varepsilon<|r|<1$. There is a map from disk to itself which is surjective from this annulus: one can do such an automorphism of a disk that $0$ is not in the annulus anymore and then use $z \mapsto z^2$. Composing this map with the projection on the first component we obtain the desired result.

Edit: actually, proof works in the assumption that $U$ and $V$ are simply-connected. I do not know whether the answer is "yes" for the covering by an annulus and a concentric circle.

(Yes). Consider a point on the boundary contained both in the closure of $U$ and $V$ and do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.

Now take the map $(x, y) \mapsto x+y$.

It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.

Edit: actually, proof works in the assumption that $U$ and $V$ contain a point on the boundary (for example if they are simply-connected and non of them is a whole $\mathbb{D}$). I do not know whether the answer is "yes" for the covering by an annulus and a concentric circle.

(Yes). Consider a point on the boundary contained both in the closure of $U$ and $V$ and do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.

Now take the map $(x, y) \mapsto x+y$.

It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.