If $\mathbf{LCA}$ denotes the category of locally compact abelian groups, then the Pontryagin dual establishes an equivalence of categories $\mathbf{LCA}^{\mathrm{op}}\to\mathbf{LCA}$. In particular, this means that sending LCA's to their duals will send limits to colimits and colimits to limits (inverse / direct limits being another term for these). Thus, the answer to your original question is yes, and your argument for proving this is correct.
Edit: originally, I had the arrows on backwards in the computation.
Regarding your example, the limit of $\dots\xrightarrow{\times p}\mathbb Z\xrightarrow{\times p}\mathbb Z$ is indeed trivial (like you mentioned), and the dual diagram $\mathbb R/\mathbb Z\to\mathbb R/\mathbb Z\to\dots$ is given by multiplication by $p$ again, like you mentioned.
However, the colimit of that diagram inside $\mathbf{LCA}$ is subtle, as it has to still be locally compact and Hausdorff. This means it can't be necessarily computed in the ambient category of general topological groups.
If we take the colimit in topological groups, the resulting group will not be Hausdorff: if we take $p=3$ for concreteness, then the element $\frac12$ (being fixed under multiplication by $p$) will be nonzero in the colimit in topological groups, but there is no open set separating it from the neutral element. For any open neighbourhood of $\frac12$ contains some element of the form $\frac n{3^m}$ for $n$, $m$ integers, which vanishes in the colimit.
However, we can correct this colimit to be a colimit in $\mathbf{LCA}$ by applying the "Hausdorffification" done in this answer, and it will (has to!) turn out that the correction will make the colimit trivial.
Best Answer
The stupid answer is that $\operatorname{Hom}_{\text{cts}}(\mathbb Z, S^1) = \operatorname{Hom}_{\mathbb Z}(\mathbb Z,S^1) = S^1$, so we are done by duality.
Here, $S^1$ denotes the unit circle $S^1 \cong \frac{\mathbb R}{2\pi\mathbb Z} \cong \mathbb R/\mathbb Z$, a.k.a. $\mathbb T$.
The real answer is that all characters of $S^1$ are of the form $z \mapsto z^n$. The reason is explained in many places, for example here: https://mathoverflow.net/questions/89504/quick-computation-of-the-pontryagin-dual-group-of-torus
In particular, you can see there that it is enough to know that $\mathbb R$ is selfdual.