I'd like to clear up something that came up in the comments. There are two natural ways to fit the finite cyclic groups together in a diagram. One is to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, m | n$ given by sending $1$ to $1$. This gives a diagram (inverse system) whose limit (inverse limit) is the profinite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$. This diagram also makes sense in the category of unital rings, since they also respect the ring structure, giving the profinite integers the structure of a commutative ring.

This is not the diagram relevant to understanding the circle group. Instead, one needs to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, n | m$ given by sending $1$ to $\frac{m}{n}$. This is the diagram relevant to understanding the cyclic groups as subgroups of their colimit (direct limit), which is, as I have said, $\mathbb{Q}/\mathbb{Z}$. And this group, in turn, "compactifies" or "completes" to the circle group in whichever way you prefer.

(These two diagrams are "dual," though, something which I learned recently when I was asked to prove on an exam that $\text{Hom}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \hat{\mathbb{Z}}$. Just observe that $\text{Hom}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \mathbb{Z}/n\mathbb{Z}$ and that contravariant Hom functors send colimits to limits!)

I'd like to clear up something that came up in the comments. There are two natural ways to fit the finite cyclic groups together in a diagram. One is to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, m | n$ given by sending $1$ to $1$. This gives a diagram (inverse system) whose limit (inverse limit) is the profinite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$. This diagram also makes sense in the category of unital rings, since they also respect the ring structure, giving the profinite integers the structure of a commutative ring.

This is not the diagram relevant to understanding the circle group. Instead, one needs to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, n | m$ given by sending $1$ to $\frac{m}{n}$. This is the diagram relevant to understanding the cyclic groups as subgroups of their colimit (direct limit), which is, as I have said, $\mathbb{Q}/\mathbb{Z}$. And this group, in turn, compactifies to the circle group in whichever way you prefer.

(These two diagrams are "dual," though, something which I learned recently when I was asked to prove on an exam that $\text{Hom}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \hat{\mathbb{Z}}$. Just observe that $\text{Hom}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \mathbb{Z}/n\mathbb{Z}$ and that contravariant Hom functors send colimits to limits!)

Edit: Let me also say something about the precise meaning of "compactification" here. A compactification of a space $T$ is an embedding $T \to X$ into a compact Hausdorff space $X$ with dense image. The embedding being considered here is the obvious one from $\mathbb{Q}/\mathbb{Z}$ to $\mathbb{R}/\mathbb{Z}$, and the fact that it has dense image is essentially what the word "completion" also means. Compactifications are not unique, but it's possible that there is a sense in which as a topological group $\mathbb{R}/\mathbb{Z}$ is the "most natural" compactification of $\mathbb{Q}/\mathbb{Z}$. But I don't know too much about topological groups.

I'd like to clear up something that came up in the comments. There are two natural ways to fit the finite cyclic groups together in a diagram. One is to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, m | n$ given by sending $1$ to $1$. This gives a diagram (inverse system) whose limit (inverse limit) is the profinite completion of $\mathbb{Z}$. This diagram also makes sense in the category of unital rings, since they also respect the ring structure, giving the profinite integers the structure of a commutative ring.

This is not the diagram relevant to understanding the circle group. Instead, one needs to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, n | m$ given by sending $1$ to $\frac{m}{n}$. This is the diagram relevant to understanding the cyclic groups as subgroups of their colimit (direct limit), which is, as I have said, $\mathbb{Q}/\mathbb{Z}$. And this group, in turn, "compactifies" or "completes" to the circle group in whichever way you prefer.

I'd like to clear up something that came up in the comments. There are two natural ways to fit the finite cyclic groups together in a diagram. One is to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, m | n$ given by sending $1$ to $1$. This gives a diagram (inverse system) whose limit (inverse limit) is the profinite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$. This diagram also makes sense in the category of unital rings, since they also respect the ring structure, giving the profinite integers the structure of a commutative ring.

This is not the diagram relevant to understanding the circle group. Instead, one needs to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, n | m$ given by sending $1$ to $\frac{m}{n}$. This is the diagram relevant to understanding the cyclic groups as subgroups of their colimit (direct limit), which is, as I have said, $\mathbb{Q}/\mathbb{Z}$. And this group, in turn, "compactifies" or "completes" to the circle group in whichever way you prefer.

(These two diagrams are "dual," though, something which I learned recently when I was asked to prove on an exam that $\text{Hom}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \hat{\mathbb{Z}}$. Just observe that $\text{Hom}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \mathbb{Z}/n\mathbb{Z}$ and that contravariant Hom functors send colimits to limits!)