It cant be proven, since it is false. [Oups, see below, the op statement may be true].

---my original reply

Hint: prime counting function π(n)

Eg: Let k = 100. Your set is thus 1,2,..., 200.

You want to partition this into 100 pairs {a_i, b_i}, such as their sum is prime.

The sum of a pair has a minimum of 3 and a max that can be 101 to 399.

But π(399) = 78. IOW, there are only 78 primes from 1 to 399. (There are less than that from 1 to 101).

But you need 100 of them, thus impossible. QED.

--end of my original reply.

BUT.... I see now...Above I assumed that the a_i + b_i gives distinct (prime) values. You did not require that.

Again, for k=100, we can chose many pairs such that their sum is 199 say: {1,198}, {2,197}...{99,100}.

This has a max of 99 pairs. In fact, you always have less than k pairs; you have at maximum k-1 pairs. Hence impossible. BUT if you consider the possibilities of pairs such as {2,197} and {2, 3} in the same set, then you can trivially generate more pairs, hence possible. And if the a+b is allowed be be greater than 2k (that is not excluded in the op, hence is permitted), then by the theorem m<p<2m, just chose that p as to generate your pairs, as other replies have pointed out.