This doesn't prove it is closed under addition, only that it is closed under addition of a vector to itself. You need to show that it is closed under addition of any two vectors:
(-y-z, y, z)+ (-b-c, b, c)= (-(y+b)-(z+c), y+b, z+c).
Or you could do it this way: If (a, b, c) and (x, y, z) are in this set then x + y + z = 0 and a+ b+ c= 0.
Then (a, b, c)+ (x, y, z)= (a+x, b+ y, c+ z) and (a+x)+ (b+y)+ (c+ z)= (a+ b+ c)+ (x+ y+ z)= 0+ 0= 0 so this sum vector satisfies the same condition and is in the subspace.
So it is closed under multiplication.
So zero vector is in the set?
Does this prove it? :S