Find all subgroups of $\mathbb Z_2\times\mathbb Z_4$ - Fraleigh p. 110 Exercise 1.11 I tried to fill in the steps but I'm still confounded by this solution. 

Any other subgroup must have order 4, since by Lagrange's Theorem, the order of any subgroup must divide 8 and:
The subgroup containing just the identity is the only group of order 1. $\color{darkred}{ \text{ 1. Why is this the only group of order 1? }}$
By cause of Fraleigh p. 100 Corollary 10.11, every group of prime order must be cyclic. $2 \in \mathbb{P}$.
The only subgroup of order 8 must be the whole group. $\color{darkred}{ \text{ 3. Why is the group itself the only subgroup of order 8? }}$
For any other subgroup of order 4, every element other than the
identity must be of order 2. Otherwise every element $\neq id$ would be cyclic and we’ve
already listed all the cyclic groups. $\color{darkred}{ \text{ 4. I don't understand this. Can someone please flesh it out? }}$
$\color{darkred}{ \text{ 7. How do you envisage and envision that there's a subgroup $\simeq V_4 \; \in \mathbb{Z_2 \times Z_4}$ ? See below.  }}$
From the above we see there are three elements of order 2: $(0,2),(1,0),(1,2)$. $\color{darkred}{ \text{ 5. How does this  induce 'hence...'? Where does ${Z_2} \times \color{blue}{<2>}$ magically crop up from? }}$
 Hence  the only other subset that could possibly be subgroup of order 4 must be $\{(0,\color{blue}0),(0,\color{blue}2),(1,\color{blue}0),(1,\color{blue}2)\} = \mathbb{Z_2} \times \color{blue}{<2>}$.
This is easily seen to be a group and completes our list. We thus have eight subgroups.
$\color{darkred}{ \text{ 6. How can you 'easily' see $\{(0,\color{blue}0),(0,\color{blue}2),(1,\color{blue}0),(1,\color{blue}2)\}$ is a subgroup? }}$
(7.) More on my question overhead. There are only two groups of order 4, up to isomorphism. By cause of Arthur's exquisite answer, I grasp the proof tries to construct $V_4$ from the elements of order $2$. This looks like a hinge to this exercise. I never envisaged this hence was confounded. 
 A: 1) If a group has order $1$, then it contains only one element (that's what "order" means). Since any other subgroup has higher order (it has to contain the identity to be a subgroup, and it has to contain some other element to be "any other subgroup"), there can be only one subgroup of order 1.
2) For any prime $p$, there is, up to isomorphism (i.e. "essentially"), exactly one group of order $p$, and that's the cyclic group of order $p$. $2$ is a prime, so it applies here as well.
3) If a subgroup has order $8$, it means it has $8$ elements. $\Bbb Z_2\times\Bbb Z_4$ has only $8$ elements, so there can only be one subgroup with $8$ elements, and that is all of the group.
4) There are, up to isomorphism, two groups of order $4$ (that's just a fact you're more or less expected to know eventually). One is the cyclic group, and another is the so-called Klein-$4$ group ($\Bbb Z_2 \times \Bbb Z_2$), and for the Klein-$4$ group it is the case that every element is its own inverse (i.e. of order $2$). All the cyclic subgroups have been listed, so if there are any order $4$ subgroups left, they have to be of the other kind.
5) He's listing all the elements of order $2$, to see if he can stitch together an example of a Klein-$4$ group. In a Klein-$4$ group there are three elements of order two, and one identity element. Since there are only three elements of order $2$, and they happen to form a subgroup together with the identity, there you are. That's the "hence".
6) There are three things that have to be checked to verify that a subset of any group is indeed a subgroup:


*

*The identity element must be present (in this case it is).

*The sum of any two elements in the subset has to be in the subset. E.g. $(0, 2)$ and $(1, 0)$ is in the subset, meaning $(0, 2) + (1, 0) = (1, 2)$ must be in the subset too. It is, and the same goes for any other sum.

*For any element in the subset, the inverse of that element has to be in the subset too. In this case, we've established that each one of those elements are its own inverse, so that's covered too.


That's how we can see that the given subset of $\Bbb Z_2\times\Bbb Z_4$ is actually a subgroup.
Added later: You didn't ask about this, but I believe it's worth mentioning: We only care about subgroups of order $1$, $2$, $4$ and $8$ because of Lagrange's theorem. Lagrange's theorem says that for any finite group, the order of any subgroup has to divide the order of the whole group. In this case, the order of the whole group is $8$, and therefore we don't have to check for subgroups of order, say, $6$ or $3$, because the only numbers that divide $8$ are $1$, $2$, $4$ and $8$. (In general there doesn't have to be a subgroup of each of these orders, but there cannot be any subgroups with other orders).
Added even later at the request from OP: As to question 7), I believe the author just went by exhaustion and said "Look, we have all the cyclic $4$-groups, what's left of the order-$4$-checking is $V_4$. Can I make a $V_4$ out of this? Why, yes, it seems like I can." Other than that, I can't really say anything about how you see or visualize subgroups and the like other than keeping up the work and get some experience in the craft. For posterity's sake, also note that all of these subgroups are of the form $G\times H$ where $G$ is a subgroup of $\Bbb Z_2$, and $H$ is a subgroup of $\Bbb Z_4$. The $V_4$, for instance, is $\Bbb Z_2 \times \langle 2 \rangle$, and the order $4$ cyclic group is $\langle 0\rangle\times \Bbb Z_4$. Remembering (and hopefully understanding) this will make it easier for you to hunt for subgroups of product groups later.
