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).
To see that a vector space is closed under addition and multiplication, you should look at the definition given in the book.
Essentially, it should boil down to something like this:
A vector space over a field $K$ is a non-empty set $V$ together with mappings $A: V \times V \to V$ and $M: K \times V \to V$ such that...
Here the two mappings $A$ and $M$ define what we want our addition and multiplication to look like, i.e. $x+y := A(x,y)$ and $\lambda x := M(\lambda, x)$.
Notice that by definition $A$ and $M$ take values (only) in $V$. This means that whenever we add to vectors $x,y$ or multiply a vector $x$ with a scalar $\lambda$, the result will still be in $V$. Hence, one says that $V$ is closed under addition and multiplication.
Best Answer
But, for any field $K$, every element of $K$ has an additive inverse. That's part of the definition of field.