$\DeclareMathOperator\colim{colim}$This is inspired by Clausen's answer.
Question: Recall that $\mathbb Z_p$ is endowed with the $p$-adic topology. Consider the countable sum $M:=\bigoplus_{n=0}^\infty\mathbb Z_p$ as a topological abelian group. One way to describe the topology is by the filtered colimit (or "inductive/direct limit topology" in the classical literature) $\colim_{N\in\mathbb N}\bigoplus_{n=0}^N\mathbb Z_p$.
- As a topological space, is $M$ first-countable? (I guess not)
- As a topological group (or more generally, a uniform space), is $M$ Cauchy-complete (i.e., every Cauchy filter converges)?
First, let's remark that $M$ is Hausdorff: for any two elements $x,y\in M$, there exists an $N\in\mathbb N$ such that $x,y$ belong to $\bigoplus_{n=0}^N\mathbb Z_p$. The result then follows from the fact that $M$ is isomorphic to $\left(\bigoplus_{n=0}^N\mathbb Z_p\right)\oplus\left(\bigoplus_{n=N+1}^\infty\mathbb Z_p\right)$. We also note that $M$ is compactly generated by definition.
We usually take the $p$-adic completion $\left(\bigoplus_{n=0}^\infty\mathbb Z\right)_p^\wedge$ in algebra, which leads to the impression that $M$ is not complete. However, after some thoughts, I find that the situation is not that simple: the opens in $M$ might be very complicated.
To see this, let $B$ denote the countable product $B:=\left(\prod_{n=0}^\infty\right)'\mathbb Z_p$ endowed with the box topology instead of the product topology. It follows, say from universal property of filtered colimit, that there exists a continuous map $M\to B$ which maps $\sum_n j_n(x_n)$ to $(x_0,x_1,\ldots)$ where $\sum_n j_n(x_n)$ is a finite sum and $j_n\colon\mathbb Z_p\to M$ is the canonical map. $B$ is not first countable, which leads me to the suspicion that neither is $M$.
Furthermore, from experience in algebra, one might mistakenly think that the series $\sum_{n=0}^\infty j_n(p^n)$ "should be" Cauchy therefore $M$ is not complete. The problem is that $j_n(p^n)$ does not converge to $0$. Indeed, $\bigoplus_{n=0}^\infty p^{n+1}\mathbb Z_p\subseteq M$ is an open neighborhood of $0$ in $M$ (which follows either from the continuity of $M\to B$, or a direct verification) to which none of $j_n(p^n)$ belongs.
Relation to condensed mathematics: it follows from Barwick-Haine Lemma 2.1.7 and Scholze's notes Lemma 1.3 that the countable direct sum $M$ coincides with the countable direct sum taken in the category of condensed abelian groups. Since $\mathbb Z_p$ is solid, so is the direct sum. It then follows from my argument that $M$ is sequentially Cauchy-complete, i.e. every Cauchy sequence in $M$ converge.
Summary: As Scholze indicated in the comment at his answer, one could replace the countable sum by arbitrary sum. One need to show that $\{U_f\}$ in that answer is indeed a basis of the topology, which follows from Tube lemma (note that $\mathbb Z_p$ is profinite therefore $p^m\mathbb Z_p$ is CHaus) and Zorn's lemma.