is $\mathbb{Z}$ a subspace of $\mathbb{R}$? My conclusion was no, because I wanted to check three things: Does it contain the $0$ vector, is it closed under addition and is it closed under scalar multiplication.
The first two ones I found to be ok, but the last one is where I disprove it. If we let c $\in$ $\mathbb{R}$ and $u \in \mathbb{Z}$ then it is obvious that we if pick some non-integer $c \in \mathbb{R}$ then $c*u \in \mathbb{R}$ and not in $\mathbb{Z}$. Therefore $\mathbb{Z}$ is not a subspace of $\mathbb{R}$.
However I read the following on a page I found online: "For example, the integers Z is contained in the real numbers R. Moreover, the addition and multiplication are compatible: The equalities 1 + 2 = 3, 2×3 = 6 hold either as integer operations or real number operations. Therefore we call Z a sub-number system of the number system R".
Is my conclusion wrong?
No, your conclusion is not wrong, since indeed $\Bbb{Z}$ is not a vector subspace of $\Bbb{R}$. The axioms of a subspace are just not satisfied, as you have shown. Of course, it is a subset of the real numbers, but this is something else.
Also $\Bbb{Q}$ is not a vector subspace of $\Bbb{R}$.