No. For example, in , if you take and , then of course they are closed, but is an additive subgroup of that is not infinite cyclic (this is because is irrational and 1 is rational). As a result, is dense in , and being a countable dense subset of the uncountable set , it cannot be closed.

But if you restrict to compact sets and , then indeed must be compact (and hence closed). This is because addition is a continuous map from to , and that is a compact subset of if both and are compact in . The first fact actually holds for general topological groups; the second fact is a consequence of the Tychonoff theorem (though of course you can prove it directly in an elementary way in this special case).