(1). I don't know what you mean by perfect, but it is correct. ${\mathbb Z}_4$ has an element of order 4, and ${\mathbb Z}_2^3$ hasn't, so no subgroup of ${\mathbb Z}_2^3$ is isomorphic to ${\mathbb Z}_4$.
(2). ${\mathbb Z}_2^3$ has 8 elements. One of them is the neutral element; all seven others have order 2.
(3). This doesn't have to be immediately clear at this point; the explanation is in the lines below. Up to isomorphism, the only groups of order 4 are ${\mathbb Z}_4$ and ${\mathbb Z}_2^2$, and in view of (1), subgroups of ${\mathbb Z}_2^3$ are not isomorphic to ${\mathbb Z}_4$. Hence the Klein group is mentioned because, whatever subgroup of order 4 you're going to find, it's going to be isomorphic to the Klein group (and not ${\mathbb Z}_4$ as just explained).
(4). This is true in general for an abelian group. The elements $x$ and $y$ are of order 2 and are distinct. Because $x$ is of order 2, $-x = x$, so $y \neq -x$, so $x + y \neq 0$. It does hold that $(x + y) + (x + y) = x + x + y + y = 0$, so $x + y$ has order 2.
(5). $\{i,x,y,x+y\}$ has 4 elements. What is left to show is that it is a subgroup, i.e., that it is closed under the group operations. For inversion: that's easy, as in ${\mathbb Z}_2^3$ every element is its own inverse anyway. For closure under addition: there are two remaining cases $x + (x + y)$ and $y + (x + y)$ and they are treated by $(\dagger)$ and $(\ddagger)$.
(6). The reasoning above says that among the 21 ways of writing down $i, x, y, x+y$ (with $x, y$ distinct and not $i$; and $x,y$ considered the same as $y,x$), there are 7 distinct sets $\{i, x, y, x + y\}$. The seven concrete distinct ways come from just starting to listing them and throwing out duplicates. Start with $\{i, a, b, a+b\}$, then $\{i, a, c, a+c\}$. After that $\{i, a, d, a+d\}$ turns out to be the same as $\{i, a, b, a+b\}$ as $a + d = b$, etc.
Edit: naming all the elements is maybe not the easiest way to list the subgroups of order 4, but is probably the most elementary way to find them. Using the symmetry inherent in ${\mathbb Z}_2^3$ the subgroups of order 4 can be described as follows. The subgroups are ${\mathbb Z}_2 \times {\mathbb Z}_2 \times \{0\}$ (plus two more of this form); the subgroup $\langle (1,1) \rangle \times {\mathbb Z}_2$ (plus two more of this form); the final one is the one generated by $(1,1,0)$ and $(0,1,1)$. They can also be seen as lines in the Fano plane. Anyway, for someone doing this particular exercise, simply listing them seems the best way to go about this.
(7). Right before (5) it is proven that they are subgroups.
The number of non-cyclic such subgroups is $0$: an abelian group of order $14$ is cyclic (by the structure theorem, CRT or some other simple arguments).
An element $(a,b)$ has order $\rm{lcm}(|a|,|b|)$. So we get the possibilities $(2,7),(7,2),(2,14),(14,2),(7,14),(14,7),(1,14),(14,1)$ and $(14,14)$. Note that in a cyclic group there are $\varphi(2)=1, \varphi(7)=6$ and $\varphi(14)=6$ elements of order $2,7$ and $14$ respectively.
Thus we get $6+6+6+6+36+36+6+6+36=144$. But to finish we need to divide by $\varphi(14)=6$, since there are that many elements of order $14$ in each such subgroup. Thus we get $24$.
Best Answer
Hint: Recall the theorem highlighted below, and note that it follows that $$\quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 6} \quad \cong \quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 2}\times \mathbb Z_{\large 3}$$
This might help to make your task a bit more clear, noting that each of $\mathbb Z_2, \; \mathbb Z_3,$ and $\,\mathbb Z_6 \cong \mathbb Z_2 \times \mathbb Z_3$ are cyclic, but $\;\mathbb Z_2 \times \mathbb Z_2,\;$ of order $\,4,\,$ is not cyclic. Indeed, there is one and only one group of order $4$, isomorphic to $\mathbb Z_2\times \mathbb Z_2$, i.e., the Klein $4$-group.
This is how we know that $\mathbb Z_6 = \mathbb Z_{2\times 3} \cong \mathbb Z_2\times \mathbb Z_3$ is cyclic, since $\gcd(2, 3) = 1.\;$
It's also why $\,\mathbb Z_2\times \mathbb Z_2 \not\cong \mathbb Z_4,\;$ and hence, is not cyclic, since $\gcd(2, 2) = 2 \neq 1$.
Good-to-know Corollary/Generalization:
The direct product $\;\displaystyle \prod_{i = 1}^n \mathbb Z_{\large m_i}\;$ is cyclic and $$\prod_{i = 1}^n \mathbb Z_{\large m_i}\quad \cong\quad \mathbb Z_{\large m_1m_2\ldots m_n}$$ if and only if the integers $m_i\,$ for $\,1 \leq i \leq n\,$ are pairwise relatively prime, that is, if and only if for any two $\,m_i, m_j,\;i\neq j,\;\gcd(m_i, m_j)=1$.