The statement of the title is true, don't listen to what some other photos might tell you. Below are two proofs.
Proof by Mathematical Induction
We induct on the number of times people have shaken hands.
Let E be the set of people who have shaken hands an even number of times.
Let O be the set of people who have shaken hands an odd number of times.
Before a handshake ever took place, O has size zero and is therefore even.
We assume the statement is true for k handshakes. We need to prove the statement for k+1 handshakes.
For the (k+1)th handshake, one of the following cases is true:
A. Two people from E shake hands (who now move to the set O).
B. Two people from O shake hands (who now move to the set E).
C. One person from O shakes hands with one person from E (they now "trade membership")
For each case, the parity of O remains the same. And since the size of O was even beforehand (by our induction hypothesis), the size of O remains even.
Seb Przd's Proof Using a Parity Argument
O and E are defined as above.
Let n be the number of times any hand has shaken another.
Let n_o be the number of times a hand from the set O has shaken another.
Let n_e be the number of times a hand from the set E has shaken another.
Note: n = n_o + n_e
Each handshake increases n by two, hence n is even. n_e is a sum of even numbers (by definition) and hence is also even. Since both n and n_e are both even, the equation above tells us that n_o is even as well. But n_o is a sum of odd numbers (by definition). The only way for n_o to be even then, is if the size of O is even.