(And besides being closed under scaling and addition, the subspace also needs to be non-empty, but you space is non-empty (it contains e.g. (0,0)).
To check that the set is closed under addition, you take two vectors in the set, add them and check if the result is again in the set. So take (x1,x2) and (y1,y2) in your set. Then x1+x2=0 and y1+y2=0. The sum of the two vectors is (x1+y1,x2+y2), and the question now remains if this is an element of the set. In order for this to be the case, it must be true that (x1+y1)+(x2+y2)=0.
This is true! Indeed, (x1+y1) + (x2+y2) = [reordering] (x1+x2) + (y1+y2) = 0 + 0 = 0.
In a similar way, you can show that the set is closed under scaling.