Let $K$ be a local field with a discrete and non-archimedean absolute value, $\mathcal{O}_K$ be its ring of integers, $\mathfrak{m}_K$ be the unique maximal ideal of $\mathcal{O}_K$ and $\mathcal{O}_K^\times$ be the groups of units of $\mathcal{O}_K$. Finally let $k := \mathcal{O}_K / \mathfrak{m}_K$ be the residue field of $K$ and $k^\times := k \setminus \{0\}$.
I am trying to prove that $\mathcal{O}_K^\times$ is isomorphic to $k^\times \times (1 + \mathfrak{m}_K)$ as topological groups. I showed (and I think it is a rather standard method) that the sequence $$1 \to (1 + \mathfrak{m}_K) \to \mathcal{O}_K^\times \to k^\times \to 1$$ (where the second arrow is the inclusion and the third is the natural projection) is exact and splits on the right. Thus, thanks to the splitting lemma, I obtained that $\mathcal{O}_K^\times \cong k^\times \times (1 + \mathfrak{m}_K)$ as groups.
However, as far as I know, the splitting lemma works only in abelian categories, and the category of topological abelian groups is not an abelian category. Hence, I cannot conclude that $\mathcal{O}_K^\times \cong k^\times \times (1 + \mathfrak{m}_K)$ also as topological groups.
Anyway, I think that at this point it should be easy with some trick, to prove that $\mathcal{O}_K^\times \cong k^\times \times (1 + \mathfrak{m}_K)$ also topological, but I have not yet figured out how to do that. I would like to show this without any computation, i.e., without without explicit the isomorphism $\mathcal{O}_K^\times \to k^\times \times (1 + \mathfrak{m}_K)$.
Thank you in advance for any suggestion.