I need to prove/disprove that $W$ is a linear subspace, and I'm not sure my approach is correct (especially the last point I'm making). Please correct me if I'm wrong.
Let $V$ be a set of vectors over $F=\mathbb{R}$, $V=\mathbb{R}^4$ and $W$ is a subgroup of $V$ such that $$W=\{(x,y,z,w)\in V|(x+y)^2=0\}$$
My solution:
For $x=0 , y=0$ we'll get $(x+y)^2=(0+0)^2=0$, thus $0\in W$.
Checking if $W$ is close under vector addition: $\forall w_1 , w_2 \in W. w_1=(x_1.-x_1,0,0) , w_2=(x_2,-x_2,0,0)|\big(x_1+x_2+(-x_1)+(-x_2)\big)^2=(0)^2=0\in W$
Checking if $W$ is close under scalar multiplication: Given $W\subseteq \mathbb{R}^4 ,\alpha\in \mathbb{R}$ we can conclude that $\forall w\in W|\alpha \times w \subseteq \mathbb{R}^4$, therefore $\forall a\in F, w\in W|\alpha \times w\in W$.
Therefore $W$ is a linear subspace of $V$.
Best Answer
You should put $v_1 = (x_1, y_1, z_1, w_1) \in W$ and $v_2 = (x_2, y_2, z_2, w_2)\in W$ and determine whether $v_1 + v_2 = (x_1+x_2, y_1+y_2, z_1 + z_2, w_1+ w_2)\in W$.
Of course, only the $x_i, y_i$ matter in determining whether the sum is in $W$, but you want to show that the sum of any two vectors in $W$ is again in $W$, not just those with $z_i, w_i = 0$. The mere existence of vectors in $W$ whose sum is again in $W$ isn't the point. To prove closure, we need to show that for any (read every) two vectors in W, their sum is again in $W$. If there exist vectors in W whose sum is NOT in $W$, then closure fails.
Now:
Is it always the case that for $v_1 + v_2 = (x_1+x_2, y_1+y_2, z_1 + z_2, w_1+ w_2)\in W$, $$(x_1+x_2)^2 + (y_1 + y_2)^2 = x_1^2 + 2x_1x_2 + x_2^2 + y_1^2 + 2y_1y_2 + y_2^2 = 0?$$ If and only if so, then do you have closure. We know that $x_1^2 + y_1^2 = 0,$ and that $x_2^2 + y_2^2 = 0$, but can you see why there may be problems with closure?
BTW: I did a double take on your proof of closure: ending with "$ 0 \in W$" makes it look like the zero vector is in W, which you showed immediately above.