Both $\{1, 3, 5, \ldots, 19\}$ and $\{11, 12, \ldots, 20\}$ are sum-free, each with $10$ elements

If $m$ is the largest element, pair $1$ to $m-1$ as $\{i, m-i\};$ each pair contributes at most one element

We can reach $10.$ The odds $\{1, 3, 5, \ldots, 19\}$ are sum-free, since two odds sum to an even. So is $\{11, 12, \ldots, 20\},$ since any two of those sum past $20.$ Each has $10$ elements. Now let $m$ be the largest element of any sum-free subset. For $1\le i<\frac{m}{2},$ at most one member of $\{i,m-i\}$ can be chosen, because the two sum to $m.$ If $m$ is even, $\frac{m}{2}$ cannot be chosen either, since it can be used twice and $\frac{m}{2}+\frac{m}{2}=m.$ Thus besides $m$ there are at most $\lfloor\frac{m-1}{2}\rfloor$ chosen elements, for a total of at most $\lfloor\frac{m-1}{2}\rfloor+1\le10.$ Therefore the greatest possible size is $10.$ Thus, C is the correct answer.