My interpretation is that the following level of detail should be given.
"Properties $2$, $3$, $7$, $8$, $9$, $10$ follow easily from the fact that they hold for $\mathbb{R}^2$." That is all one needs to say about these. But it must be said.
We now deal with the remaining axioms.
What you wrote for $1$ is not bad, but to be totally explicit it should have ended with $(x_1,2x_1)+(x_2,2x_2)=(x_1+x_2, 2x_1+2x_2)=(x_1+x_2,2(x_1+x_2)).$ So the sum of our two vectors is of the shape $(u,2u)$, and therefore is in our set.
It is handy to give our set a name, such as $U$, because we will be referring to it often.
What you ended that part of the argument with, namely $\in\mathbb{R}^2$ could be a problem. Of course the thing is in $\mathbb{R}^2$, but that's not what needs to be shown. The fact you wrote it down could be interpreted as indicating confusion about what it means to be a subspace.
Next, for $4$, we need to show that the $0$ vector is in $U$. So all we need to do is to show it has the right shape. This is easy, $(0,0)=(0,2(0))$. The properties of the zero vector do not need proof, they are inherited from $\mathbb{R}^2$.
For $5$, we need to show that the ordinary additive inverse of an element of $U$ is in $U$. So look at the additive inverse of $(x,2x)$. It is $(-x,-2x)$, which is equal to $(-x,2(-x))$, so it is in $U$. Everything else in $5$ is inherited.
Finally, $6$. We need to show that if $(x,2x)\in U$ then $c(x,2x) \in U$. This is just as easy as all the others: $c(x,2x)=(cx,c(2x))=(cx,2(cx))$ and obviously (love that word) $(cx,2(cx))\in U$.
Your proposed handling of scalar multiplication was not good. What needs to be shown is that $U$ is closed under scalar multiplication, and there was no apparent attempt to do that.
As soon as what actually needs to be verified is clear, the verifications themselves are unchallenging. What is being tested in the problem is whether you fully know the meaning of subspace.