Informal proof that it's impossible. (This is not rigorous, so I may just be missing a solution, but oh well.) See attached image. It should be clear that you can move the nodes around without affecting anything, since we're not limited in the length or direction of our lines.
Drawing lines from A to X,Y,Z and from B to X,Y,Z will always divide the plane into three areas: two 4-sided polygons which share two sides (blue and green in my attachment), and an area outside those polygons (white).
For proof that it will always form these areas: Take A and B, and any two of X,Y,Z, let's say X and Y, though it doesn't matter. Connecting them as per the original problem will always form a quadrilateral. This should be obvious; if you connect four points with four lines, two lines touching each point, you form a 4-sided polygon AXBY. Now you have one leftover node (Z). You can either place it inside or outside the polygon, nowhere else. If you place it inside, the two lines from A and B will split the polygon into two pieces, which leaves you with two 4-sided polygons sharing two sides. If you place the node outside, the two lines from A and B will form a second polygon which borders the first (sharing two sides). Either way, you end up with two enclosed 4-sided polygons sharing two sides, as per my attachment.
If C is enclosed in AZBY (blue) it can't reach X. If C is enclosed in AYBX (green) it can't reach Z. If C is outside them both (white) it can't reach Y. Forming the two 4-sided polygons from nodes A,B,X,Y,Z or A,C,X,Y,Z or B,C,X,Y,Z all reduce to the above scenario. There are no other possibilities; therefore the original problem is impossible.
EDIT: Clearly if you can show any way that you can connect two of A,B,C with three of X,Y,Z without forming two 4-sided polygons that share two edges, then what I've posted will be disproved.
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