I want to show that the $p$-adic integers is torsion-free, but I am not sure if my proof is correct.
Here I'm thinking about the inverse limit definition of $p$-adic integers. Thus I want to show that there does not exists an element that looks like a sequence $\{a_i\}_{i \ge 1}$ such that $a_i \in \mathbb{Z}/p\mathbb{Z}$ and $a_i \equiv a_j$ for $i \le j$.
Let $n < \infty$ such that $n \cdot (a_1, a_2, a_3, \ldots) = 0 \in \mathbb{Z}_p$. This means that $a_i n = p^i$ for all $i\ge 1$. Thus $n = 1$ or $p$.
If $n = 1$, then $a_i = p^i$ for all $i$ and hence $(a_1, a_2, a_3, \ldots) = 0 \in \mathbb{Z}_p$.
If $n = p$, then $a_i p = p^i$ implies that $a_i = p^{i-1}$ for all $i$, but $a_i \not\equiv a_j$ for $i \le j$ so this element is not in $\mathbb{Z}_{p}$.
Is this the correct way to see it?
This is not a complete argument to show "torsion-free-ness," as I think you only want a comment on the part of the argument that you have given.
Your $n=1$ case is fine, although it might be a bit neater to write $a_i=0\in {\mathbb Z}/p^i{\mathbb Z}$ - but of course $0=p^i \in {\mathbb Z}/p^i{\mathbb Z}$.
But your $n=p$ case/conclusion is not correct... One has $a_i = p^{i-1}b_i$, for some $b_i$, for $i\ge 2$. On the other hand, relabeling, one has $$a_{j+1} = p^j b_{j+1} \equiv a_j,$$ for all $j\ge 1$. Therefore $a_j=0\in {\mathbb Z}/p^j{\mathbb Z}$.