Is the following statement obvious and, if not, how would you show it is true?

To me the statement falls just short of being obvious. This is the best explanation I can think of, which seems more complicated than it ought to be.

Designate the committees as Committee 1 and Committee 2, and designate the members of Committee 1 as A and B. Since Committee 2 is not identical to Committee 1 and the same size, it must have at least 1 member who is different from A and B, who we designate as C. There are only two possibilities for the second member of Committee 2, A or B. In either case, there will be exactly one person who is on both committees.

Here is another approach. Since each committee has two people, they must overlap since otherwise they would have a total of 2×2=4 members, but there are only a total of 3 people. So the committees have at least one member in common. If a committee shared more than 1 person with the other, they would have 2 members in common, which would make them identical, so they must have exactly one person in common.

If you label the three people A, B and C then look at all possible committees they could be on; AB, BC and AC then it is clear that whichever committee you look at one member of it will also be a member of another committee.

Each committee would have exactly 1.5 people, mathematically. Since you can’t have half a person, you would have to have 1 person serving on 2 committees.

The people are A, B and C.

The Committee #1 has A and B on it.

The Committee #2 has B and C on it.

It is obvious.

If it isn’t to someone, I’d ask them to try examples, until they satisfy themselves.

It is obvious. I’m not sure how I would ‘show’ it. I haven’t done proofs since college and wasn’t terribly good at them even then. I was good at knowing whether a thing was true or not, but terrible at proofing it out.