Identify Subspaces I read a post and have a better understanding of what subspace is

Let $V$ be the set of vectors $\langle x,y,z\rangle\in\Bbb R^3$ such that $x-y+2z=0$ and $x+y+4z=0$; then $V$ is a subspace of $\Bbb R^3$.

In order to show this, you must show three things:
(1) $V$ is non-empty.
(2) $V$ is closed under vector addition: if $\vec u,\vec v\in V$ then $\vec u+\vec v\in V$.
(3) $V$ is closed under scalar multiplication: if $\vec v\in V$ and $\alpha\in\Bbb R$, then $\alpha\vec v\in V$.
I don't understand what "V is closed" mean in this definition
There's an example from the textbook Introduction to Linear Algebra by Gilbert Strang:

Find the subsets of $\Bbb R^3$ that are actually subspaces.
(a) The plane of vectors $(b_1, b_2, b_3)$ with $b_1 = b_2$
(b) The plane of vectors with $b_1 = 1$
(c) The vectors with $b_1b_2b_3 = 0$
(d) All linear combinations of $v = (1, 4, 0)$ and $w = (2, 2, 2)$
(e) All vectors that satisfy $b_1 + b_2 + b_3 = 0$
(f) All vectors with $b_1 \leqslant b_2 \leqslant b_3$

The correct answer is (a), (d) and (e).
For (a), we know $(b_1, b_2, b_3)$ belongs to $\Bbb R^3$, so definition (1) is satisfied. But for definition (2) and (3), since $b_1, b_2, b_3$ $\in  \Bbb R$, isn't it $b_1 + b_2 + b_3$ $\in  \Bbb R$ as well?
For (b), it only says The plane of vectors but it doesn't say how many b's are there. so we cannot garentee is its $\Bbb R^3$. Is this the correct reasoning?
For (c), same reason for (a), $b_1 * b_2 * b_3 \in  \Bbb R$, which is not $\Bbb R^3$. So (c) is false
For (d), since multiplication and addition will make $\Bbb R^3$, so (d) is correct.
For (e), even though it says "All vectors that satisfy $b_1 + b_2 + b_3 = 0",$ the multiplication is already $\Bbb R$... Why is it true?
For (f), same question as (e)..
 A: All "is closed under an operation" means is that performing the operation on members of that set produces new members of that set. (See here for the Wikipedia article.)
In this specific case, $V$ is closed under addition if and only if, for any vectors $v_1, v_2$ in V, if we perform the operation of addition with them, the resulting vector $v_1 + v_2$ is again in $V$.
Likewise, $V$ is closed under scalar multiplication if and only if, for any vector $v$ in $V$, if we multiply it by an arbitrary scalar $\lambda$, then the resulting vector $\lambda v$ is again in $V$.
For (a) and (c), you seem to be confusing a vector $b = (b_1, b_2, b_3)$ with its components, $b_1, b_2, b_3$. It is a subtle difference, but an important one. If we add two vectors in the plane, we add their components element-wise, e.g. if we have $b = (b_1, b_2, b_3), c = (c_1, c_2, c_3)$, then $b+c = (b_1 + c_1, b_2 + c_2, b_3 + c_3)$.
For (b), the reasoning is unfortunately not correct. The set is non-empty, since the vector $b = (1, 0, 0)$ is in it.
The reasoning for (d) is unfortunately also incorrect. $\mathbb{R}^3$ is a three-dimensional vector space, which can be spanned by only at least three vectors. But the space in question is spanned by only two vectors.
(e)-(f) I don't understand what you mean by the "multiplication is already $\mathbb{R}$.
A: 
I don't understand what "V is closed" mean in this definition

The meaning of "closed" is precisely stated afterwards in symbols:

(2) $V$ is closed under vector addition: if $\vec u,\vec v\in V$ then $\vec u+\vec v\in V$.
  (3) $V$ is closed under scalar multiplication: if $\vec v\in V$ and $\alpha\in\Bbb R$, then $\alpha\vec v\in V$.

In words, slightly informal: (all) sums of elements of the set have to stay in the set and (all) scalar multiples of elements in the set have to stay in the set.
To show that a subset isn't a subspace, it suffices to find a counter-example to the conditions above.

(b) The plane of vectors with $b_1 = 1$

Then, for example, the element $(\color{blue}{1},1,1)$ is in the subset (because $b_1=\color{blue}{1}$) but the scalar multiple $2\cdot (1,1,1) = (\color{red}{2},2,2)$ isn't (because $b_1=\color{red}{2} \ne 1$), so not all scalar multiples stay in the subset, hence it is not a linear subspace.
Or, with the sum condition: $(\color{blue}{1},0,1)$ and $(\color{blue}{1},1,0)$ are in the subset, but their sum $(1,0,1)+(1,1,0) = (\color{red}{2},1,1)$ isn't.

If you understand this, try finding simple counter-examples for (c), (d) and (f) as well.
When a subset is in fact a subspace, you cannot prove this by using one or more examples. In that case you'll need to show that the conditions (non-empty and closed under addition and scalar multiplication) hold in general.
Your textbook probably gives an example of how to do this: check if the zero vector is in the subset (non-empty) and then pick some arbitrary element(s) to verify the other two conditions.

Additions after comment

For (c), I think it is incorrect because even though $b_1b_2b_3=0$, it is possible that $(b_1, b_2, b_3) = (-1, 1, 0)$. So $3 \cdot (−1,1,0)=(−3,3,0)$, and $b_1$ and $b_2$ do not stay in the subset. 

No, that's not right since also $(−3,3,0)$ is in this. You have to check whether $b_1b_2b_3=0$ and $-3 \cdot 3 \cdot 0 = 0$. Notice that the requirement $b_1b_2b_3=0$ means that at least one coordinate is $0$.
Hint: find two elements with this property, but such that their sum doesn't have the property.

For (d), I don't quite understand what "all linear combination" means.

All the linear combinations of $v=(1,4,0)$ and $w=(2,2,2)$ are all the elements of the form $a \cdot (1,4,0)+b \cdot (2,2,2)$ for any real numbers $a$ and $b$.

For (e), I thought it is similar to (c), but why is (e) correct?

The counter-example to (c), see the hint above, does not work for (e). 
Either show it in general (take two arbitrary elements et cetera) or, alternatively: note that if $b_1+b_2+b_3=0$, then any element $(b_1, b_2, b_3)$ is of the following form:
$$(b_1, b_2, \color{blue}{b_3})=(b_1, b_2, \color{blue}{-b_1-b_2})=b_1 \cdot (1,0,-1) + b_2 \cdot (0,1,-1)$$so you're looking at all the linear combinations of $(1,0,-1)$ and $(0,1,-1)$ and the result is similar to part (d). Or in yet another way: geometrically $b_1+b_2+b_3=0$ represents a plane through the origin in the coordinates $b_1$, $b_2$, and $b_3$ and planes through the origin are subspaces.
