So, I was taught that $\mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}$.
In general, we're often told that $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$, but, alas, none of these containments are true.
For instance, $\mathbb{Z} := \mathbb{N}^2/R$, where $R$ is the equivalence relation defined by
$$R = \{ ((a,b),(c,d)) \in \mathbb{N}^2\times\mathbb{N}^2 \mid a+d = b+c\}$$ (see here for full details). With this definition, it is clear that $\mathbb{N}$ is not a subset $\mathbb{Z}$. However, one can identify every natural number $n$ with the equivalence class $[(n,0)]$.
As far as the original question, it is not the case that $\mathbb{R} \subset \mathbb{C}$. However, the mapping $x\in \mathbb{R} \longmapsto x+0i \in \mathbb{C}$ is a ring homomorphism (and, hence, an embedding) and an isometry (notice that $\vert{x-y}\vert = \begin{Vmatrix} (x + 0i)-(y+0i)\end{Vmatrix}$). Thus, the real axis is a subfield of $\mathbb{C}$ that is isomorphic to $\mathbb{R}$.