My teacher said to use Bertrand's postulate and I have tried this for so long and I seem to go nowhere. Help would be appreciated.
EDIT: Here's what I've done in my proof so far (I need help finalizing case 2)
First note that for $p_0=13$, we can express all integers $7 \leq n \leq 2p_0=26$ as a sum of distinct primes less than or equal to $13$. Now, we will prove that we can construct these sums indefinitely. Assume we know some prime $p'$ exists such that every integer $7\leq n \leq 2p'$ can be expressed as a sum of distinct primes less than or equal to $p'$. Then, by Bertrand's postulate, there exists a prime $p$ such that $p'<p<2p'$. We claim now that every integer $7\leq n \leq 2p$ can be written as a sum of distinct primes less than or equal to $p$. Consider the two following cases
Case 1: $2p'-p\geq 7$, hence $2p'\geq p+7$ so the terms $p,p+1,\dots,p+7$ are less than or equal to $2p'$ which means they can be written as a sum of distinct primes $\leq p'$ by hypothesis. It is left to check whether the terms $p+8,p+9,\dots, 2p$ satisfy our claim. Note if we subtract $p$ from every term in the arithmetic progression above, it becomes $8,9,\dots, p<2p'$ which shows that each term can be written as the sum of $p$ plus some other distinct primes less than or equal to $p'<p$ by hypothesis.
Case 2: $2p'-p\leq 6$, hence $2p'\leq p+6$. Here we can see all terms $p+7,\dots, 2p$ satisfy our claim along with $p+2,p+3$ and $p+5$ by a similar argument as in Case 1. I'm not sure how to deal with $p+1,p+4,$ and $p+6$ though.
EDIT: Oh, since any prime $p \geq 13$ is odd, then the only possible values for $2p'$ in Case 2 are $p+1,p+3$ and $p+5$, so I don't have to worry about the other cases. I think I'm done!
EDIT: Nope, I still need to deal with $p+4$ and $p+6$.