Verify whether the following set is a subspace of the vector space Verify whether the following set is a subspace of the vector space taken into consideration:
$\{(x,y,z) \mid x=y+2z\}$, in $\mathbb{R}^3$ over $\mathbb{R}$.
Is my solution ok?

I'm checking if it can be zero vector: $x-y-2z=0$ so if $x=0$, $y=0$ and $z=0$ it is ok.
Vector addition: It is ok for $(6,2,2)$. Is it enough?
Now I'm checking scalar multiplication: For example $-1(6,2,2)$ it's still ok? $-6=-2-2\cdot2$

Do I need to prove it in some other way?
 A: I am mostly just repeating what JMoravitz has said in the comments, but I hope that the extra length allowed in a full answer will help clarify the issue:
First, let's put a label on that set, so we can reference it more easily:
Let $V = \{(x, y, z) \in \Bbb R^3 \mid x = y + 2 z\}$. To show that $V$ is a subspace of $\Bbb R^3$, we need to show that three things are true:


*

*$\mathbf 0 \ = (0,0,0) \in V$

*For every $\mathbf v_1, \mathbf v_2 \in V$ we also have $\mathbf v_1 + \mathbf v_2 \in V$

*For every $\mathbf v \in V$ and $\alpha \in \Bbb R$, we also have $\alpha\mathbf v \in V$.


As JMoravitz as pointed out, these three conditions together are equivalent to showing that for all $\mathbf v_1, \mathbf v_2 \in V$ and $\alpha, \beta \in \Bbb R$, we also have $\alpha v_1 + \beta v_2 \in V$. But since you are a novice at this, I think it might be wiser to stick with the three invididual conditions for now. 
So, let us start with (1). How do you show that $\mathbf 0 \in V$? Well, by definition $(x, y, z) \in V$ if and only if $x = y + 2z$. So to show that $\mathbf 0 = (0,0,0) \in V$, we just have to note that $(0) = (0) + 2(0)$.
For (2), I am not sure what you mean by "it is okay for $(6,2,2)$". Vector addition is about the sum of two vectors, but you have only given one. So you cannot have shown that "it is okay" for that vector. But in any case, to show the $V$ is a vector space, you must show that for EVERY pair of vectors in $V$ that their sum is also in $V$. One case does not prove this. 
So instead, we start with two arbitrary vectors $\mathbf v_1 = (x_1, y_1, z_1)$ and $\mathbf v_2 = (x_2, y_2, z_2)$ in $V$. We must show that $\mathbf v_1 + \mathbf v_2 \in V$. Since $\mathbf v_1, \mathbf v_2 \in V$, we know that
$$x_1 = y_1 + 2z_1\\x_2 = y_2 + 2z_2$$
Now, $\mathbf v_1 + \mathbf v_2 = (x_1 + x_2, y_1 + y_2, z_1 + z_2)$. To show that it is in $V$, we must show that
$$(x_1+x_2) = (y_1+y_2) + 2(z_1+z_2)$$
Can you figure out how to use the facts that
$$x_1 = y_1 + 2z_1$$
and 
$$x_2 = y_2 + 2z_2$$
to prove $$(x_1+x_2) = (y_1+y_2) + 2(z_1+z_2)\text{?}$$
For (3), the approach is similar. Let $\mathbf v = (x, y, z) \in V$. Then we know that $x = y + 2z$ Now for $\alpha \in \Bbb R$, by definition, $\alpha \mathbf v = (\alpha x, \alpha y, \alpha z)$. To prove (3), you must use the fact that $$x = y + 2z$$ to prove that $$(\alpha x) = (\alpha y) + 2(\alpha z)$$
