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If you allow the open sets $X$, $U$ and $V$ to be disconnected, you get a counter-example by taking a countable example and changing points to pairwise disjoint balls.

Edit 2: Here is a modified version of Joel's example, which was later deleted. Define $U$ to be the open set inside the green rectangle and $V$ to be the open set inside the red rectangle. Just for notation, suppose that the left-bottom corner is $(0,0)$ and that the width of the "house" is $1$.

Define $f \colon U \to U$ be a mapping $$(x,y) \mapsto \left(\frac12x,y\right)$$ on the part of $U$ which does not include the "chimney", i.e. the top left box. Correspondingly, the part of the mapping $g \colon V \to V$ which maps everything but the chimney is defined as $$(x,y) \mapsto \left(\frac12(x+1),y\right).$$ The chimneys are then mapped with homoemorphisms to the missing parts of $U$ and $V$. Now the set $U \cap V$ is the middle part of the house. The set inside the yellow rectangle has two pre-images and has positive area.

Remark: This example does not differ that much from the first one which I posted. Essentially we just extend the mappings defined on the balls slightly (and rearrange the balls if this is not possible otherwise) to make the sets $U$ and $V$ connected.

If you allow the open sets $X$, $U$ and $V$ to be disconnected, you get a counter-example by taking a countable example and changing points to pairwise disjoint balls.

Edit 2: Here is a modified version of Joel's example, which was later deleted. Define $U$ to be the open set inside the green rectangle and $V$ to be the open set inside the red rectangle. Just for notation, suppose that the left-bottom corner is $(0,0)$ and that the width of the "house" is $1$.

Define $f \colon U \to U$ be a mapping $$(x,y) \mapsto \left(\frac12x,y\right)$$ on the part of $U$ which does not include the "chimney", i.e. the top left box. Correspondingly, the part of the mapping $g \colon V \to V$ which maps everything but the chimney is defined as $$(x,y) \mapsto \left(\frac12(x+1),y\right).$$ The chimneys are then mapped with homoemorphisms to the missing parts of $U$ and $V$. Now the set $U \cap V$ is the middle part of the house. The set inside the yellow rectangle has two pre-images and has positive area.

Remark: This example does not differ that much from the first one which I posted. Essentially we just extend the mappings defined on the balls slightly (and rearrange the balls if this is not possible otherwise) to make the sets $U$ and $V$ connected.

If you allow the open sets $X$, $U$ and $V$ to be disconnected, you get a counter-example by taking a countable example and changing points to pairwise disjoint balls.

Edit: Because I still don't see how Joel's example works, I decided to modify it by drilling infinitely many holes (windows) to the house. Still the same shrinking by affine mapping of everything but the chimneys is performed to the house, but now the chimneys are mapped so that they fill in the space above the top window. The set $U$ is everything left from the blue line and $V$ is symmetrically from the other side.

If you allow the open sets $X$, $U$ and $V$ to be disconnected, you get a counter-example by taking a countable example and changing points to pairwise disjoint balls.

Edit 2: Here is a modified version of Joel's example, which was later deleted. Define $U$ to be the open set inside the green rectangle and $V$ to be the open set inside the red rectangle. Just for notation, suppose that the left-bottom corner is $(0,0)$ and that the width of the "house" is $1$.

Define $f \colon U \to U$ be a mapping $$(x,y) \mapsto \left(\frac12x,y\right)$$ on the part of $U$ which does not include the "chimney", i.e. the top left box. Correspondingly, the part of the mapping $g \colon V \to V$ which maps everything but the chimney is defined as $$(x,y) \mapsto \left(\frac12(x+1),y\right).$$ The chimneys are then mapped with homoemorphisms to the missing parts of $U$ and $V$. Now the set $U \cap V$ is the middle part of the house. The set inside the yellow rectangle has two pre-images and has positive area.

Remark: This example does not differ that much from the first one which I posted. Essentially we just extend the mappings defined on the balls slightly (and rearrange the balls if this is not possible otherwise) to make the sets $U$ and $V$ connected.

If you allow the open sets $X$, $U$ and $V$ to be disconnected, you get a counter-example by taking a countable example and changing points to pairwise disjoint balls.

If you allow the open sets $X$, $U$ and $V$ to be disconnected, you get a counter-example by taking a countable example and changing points to pairwise disjoint balls.

Edit: Because I still don't see how Joel's example works, I decided to modify it by drilling infinitely many holes (windows) to the house. Still the same shrinking by affine mapping of everything but the chimneys is performed to the house, but now the chimneys are mapped so that they fill in the space above the top window. The set $U$ is everything left from the blue line and $V$ is symmetrically from the other side.