List all the cyclic subgroups of $\langle\mathbb Z_{10}, +\rangle$ I am trying to get the hang of it, so my work is very belabored. Please, see if any of that is correct. 
Counterexamples: 
$\mathbb Z_2 = \{0, 1\}$ is not subgroup since $1 + 1 = 2$ is outside $\mathbb Z_2.$
$\mathbb Z_3 = \{0, 1, 2\}$ is not a subgroup since $3 + 1 = 4$ is outside $\mathbb Z_5.$
$\mathbb Z_4 = \{0, 1, 2, 3\}$ is not a subgroup since $3 + 1 = 4$ is outside $\mathbb Z_4.$
$\mathbb Z_5 = \{0, 1, 2, 3, 4\}$ is not a subgroup since $4 + 1 = 5$ is outside $\mathbb Z_3.$
$\mathbb Z_6 = \{0, 1, 2, 3, 4, 5\}$ is not a subgroup since $5 + 1 = 6$ is outside $\mathbb Z_6.$
$\mathbb Z_7 = \{0, 1, 2, 3, 4, 5, 6\}$ is not a subgroup since $6 + 2 = 8$ is outside $\mathbb Z_7.$
$\mathbb Z_8 = \{0, 1, 2, 3, 4, 5, 6, 7\}$ is not a subgroup since $5 + 3 = 8$ is outside $\mathbb Z_8.$
$\mathbb Z_9 = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$ is not a subgroup since $7 + 2 = 9$ is outside $\mathbb Z_9.$

$\mathbb Z_1 = \{0\}$.
$0 + 0 = 0$, so $0 \in \mathbb Z_1$ and $0$ is the identity and its own inverse, so $\mathbb Z_1$ is a subgroup of $\mathbb Z_1$.
$\mathbb Z_{10} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$
Let $c \neq 0 \in \mathbb Z_{10}.$ Then $c + 0 = c, \forall c \in \mathbb Z_{10}.$ So, $0$ is the identity. $c + d = 0 \to d = -c,$ so $-c$ is the inverse. The sum of any two elements in $Z_{10}$ is still in $Z_{10}.$ So $\mathbb Z_{10}$ is a subgroup of $\mathbb Z_{10}$. 

Cyclic subgroups are those generated by a single element. Let's look at $H = \langle 2 \rangle$ as an example. This means the subgroup generated by $2$. So, we start off with $2$ in $H$, then do the only thing we can: add $2 + 2 = 4$. So, just by having $2$, we were able to reach $4$. Now we know that $2$ and $4$ are both in $H$. We already added $2 + 2$, so let's try $2 + 4 = 6$. Now we know $6$ is in $H$ as well. Next would come $6 + 2 = 8$, then finally $8 + 2 = 10 \equiv 0 \mod 10$. Now if we try $0 + 2 = 2$, we've landed back at something we started with, $2$. Thus we say that $\langle 2 \rangle = \{0,2,4,6,8\}$ is a subgroup of $\mathbb{Z}_{10}$, and it's cyclic because it's generated by a single element, in this case $2$.
I think you've chosen a bad direction to find these - for example, I see $\{0, 1, 2\}$ listed, but you know that $\mathbb{Z}_{10} = \langle 1 \rangle$, so any subgroup containing $1$ must necessarily be the whole group.
Since cyclic subgroups are those generated by a single element, why not pick each element of $\mathbb{Z}_{10}$ and see what subgroup it generates, by adding that element to itself until you get back to where you started? For example, $\langle 2 \rangle = \{0, 2, 4, 6, 8\}$.
