Prove that every proper subgroup of $C_{p^\infty}$ is finite cyclic. (Here $C_{p^\infty}$ is the prufer p-group.)
$C_{p^\infty} = \langle x_1,x_2,.... | px_{i+1}=x_i, px_1=1 \rangle$
Proof: Let $H$ be a proper subgroup of $C_{p^\infty}$. Then there exists a minimum $j$ such that $H \leq \langle x_j \rangle$. Then $|H|$ divides $p^j$, and $H$ does NOT divide $p^{j-1}$. Therefore $|H| = p^j$ and it follows that $H = \langle x_j \rangle.$
Is this correct?
The idea of the proof is correct, but it is missing detail.
Firstly, it is not obvious that $H$ proper implies that $H \leq\left<x_j\right>$. Instead, consider the set $I = \{i, x_i \in H\}$. Since $H$ is proper, it is different from $\mathbb{N}^*$. By the propertiers of $C_{p^\infty}$, if $i \in I$, then $x_{i-k} = p^{k}x_i$ is also in $H$, so $i-k$ is in $I$ for all $k \leq i-1$, so I must be of the form $\left\{1,\ldots, j\right\}$.
EDIT:
This shows that $\left<x_j\right> \leq H$. To show the converse, that is $H \leq \left<x_j\right>$, you can proceed as follows, assume $H$ is not contained in $\left<x_j\right>$. Take an element $h \in H$ such that $h \not \in \left<x_j\right>$. By the definition of $C_{p^\infty}$, $h$ must be a finite product of generators. Write $h = \sum\limits_{i=1}^me_ix_i$ (the group is abelian, so I use additive notation). With $e_i \geq 0$ for all $i$, and $e_m > 1$ which is not a power of $p$. Since $h \not\in \left<x_j\right>$, $m > j$ (otherwise it would be a sum of elements of $\left<x_j\right>$.
Then, $ph = \sum\limits_{i=2}^me_ix_{i-1}$ (since $px_1 = 0$), wich also writes $ph = \sum\limits_{i=1}^{m-1} e_{i+1}x_{i}$. By iterating, one has $p^{m-j+1}h = \sum\limits_{i=1}^{j+1}e_{i+m-j+1}x_{i}$. So an element of $H$ is of the form $ex_{j+1}h'$, where $h' \in \left<x_j\right>$. Hence $ex_{j+1} \in H$, since $e$ is not e a power of $p$, it is invertible mod $p^{j+1}$, so that there exists an integer $f$ such that $fex_{j+1} = x_{j+1}$, hence $x_{j+1} \in H$, which contradicts the minimality of $j$.
The rest of the proof is then as you wrote.