I would like to say that $\Bbb Q / \Bbb Z$ is not discrete (when $\Bbb Q$ has euclidean topology), since $\Bbb Z \subset \Bbb Q$ is not open. But OTOH we have $$\Bbb Q / \Bbb Z \cong \bigoplus_p \Bbb Q_p / \Bbb Z_p$$ which is a direct sum of discrete groups, so it should be a discrete group. Maybe the issue is that the above isomorphism is only as abstract groups, but not as topological groups.
Could anyone confirm/elaborate on this?
A non-trival group $G$ can always be endowed with various distinct topologies making it a topological group. Two "universal choices" are the discrete topology and the indiscrete topology. See What is, exactly, a discrete group?.
However, you consider $G = \Bbb Q / \Bbb Z$ and emphasize that $\Bbb Q$ has the Euclidean topology. In that case the only reasonable topology on $G$ will be the quotient topology which is certainly not discrete.
The isomorphism $\Bbb Q / \Bbb Z \cong \bigoplus_p \Bbb Q_p / \Bbb Z_p$ is therefore only an algebraic isomorphism, not an isomorphism of topological groups.
Edited: I just considered which topology is given to an infinite sum $\bigoplus_{\alpha \in A} G_\alpha$ of abelian topological groups. There are various approaches, see for example
Higgins, P. J. "Coproducts of topological Abelian groups." Journal of Algebra 44.1 (1977): 152-159.
https://core.ac.uk/download/pdf/82771298.pdf
Chasco, M. J., and X. Dominguez. "Topologies on the direct sum of topological abelian groups." Topology and its Applications 133.3 (2003): 209-223.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.506.7942&rep=rep1&type=pdf
In my opinion the conclusion is that an infinite sum of discrete abelian topological groups is not given the discrete topology. Whether one of the "reasonable" topologies on $\bigoplus_p \Bbb Q_p / \Bbb Z_p$ makes it isomorphic as a topological group to $\Bbb Q / \Bbb Z$ is not known to me.