Your proof is fine. Also, to prove that a subset $S$ is not subspace, it suffices to show that at least one of the conditions ($0 \in S$, closure under addition and scalar multiplication) fails. So as soon as you show that $0 \notin S$, your proof is complete.
To show a subset is a subspace, you need to show three things:
- Show it is closed under addition.
- Show it is closed under scalar multiplication.
- Show that the vector $0$ is in the subset.
To show 1, as you said, let $w_{1} = (a_{1}, b_{1}, c_{1})$ and $w_{2} = (a_{2}, b_{2}, c_{2})$. Suppose $w_{1}$ and $w_{2}$ are in our subset. Then they must satisfy $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$. We need to check that $w_{1} + w_{2}$ is in our subset, that is, we need to check that if
$$w_{1} + w_{2} = (a_{1} + a_{2}, b_{1} + b_{2},c_{1} + c_{2})$$
then it satisfies that the first coordinate is greater than or equal to the second coordinate. Is $a_{1} + a_{2} \geq b_{1} + b_{2}$? Yes, by adding the two inequalities $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$ together. So $w_{1} + w_{2}$ is in our subset.
For 2, we need to show that if $\alpha$ is any scalar, then if $w_{1} = (a_{1}, b_{1}, c_{1})$ is in our subset, so is $\alpha w_{1}$. But $\alpha w_{1} = (\alpha a_{1}, \alpha b_{1}, \alpha c_{1})$. We know since $w_{1}$ is in our subset that $a_{1} \geq b_{1}$. We need to check that for any $\alpha$, $\alpha a_{1} \geq \alpha b_{1}$. This inequality is definitely true if $\alpha \geq 0$, but we need it to be true for all $\alpha$, and it's not, because if $\alpha$ is negative, then multiplying the inequality $a_{1} \geq b_{1}$ on both sides by a negative number makes the inequality flip. So we would get $\alpha a_{1} \leq \alpha b_{1}$ if $\alpha < 0$, which means the inequality doesn't hold. Thus, if $\alpha < 0$, then $\alpha w_{1}$ is not in our subset because it doesn't satisfy $\alpha a_{1} \geq \alpha b_{1}$.
So our subset is not a subspace because it doesn't satisfy 2 (it is not closed under scalar multiplication, because any negative scalar would cause this problem).
Best Answer
To begin with, the statement $W = \text{span}(W)$ means every linear combination of elements of $W$ is also in $W.$ Your interpretation is also a part of two sets being equal (that would be $W \subseteq \text{span}(W)$) but this part is trivially true because you can simply say each vector in $W$ is equal to itself, which is a linear combination of vectors from $W.$
As for the closures, the closures under addition and multiplication are two of the axioms which any set must fulfill in order to be a vector space. A subspace of a vector space is also a vector space, so it must fulfill these axioms.
Closure under addition means that adding any two members of the vector space will always give you another member of the vector space. A good example would be the set of integers, since whenever I add two integers I always get another integer. An example of a set not closed on addition is odd whole numbers: $3$ and $5$ are odd and $3 + 5 = 8,$ but $8$ is not odd, so the set is not closed under addition.
Closure under scalar multiplication is similar, except now when we multiply a member of a set by a scalar from some other field, we get another member of the set. As an example, let's let the candidate space be irrational numbers and the field of scalars can be the integers. The irrational numbers are closed under multiplication by an integer because if I multiply any irrational number by an integer, I must get another irrational number, since the product being representable by a rational fraction would give me a way to represent the original number as a rational fraction. An example of a non-closed set could be the set of odd integers under multiplication by the integers: $3$ is odd and $2 \cdot 3 = 6,$ but $6$ is not odd.
Notice also that if your field contains a $0,$ then your closure under multiplication also guarantees that the zero vector is in your subspace.
Now do you see how you can proceed?