Let's stick to $\mathbb{R}^3$ for the sake of simplicity. Say I've got a sphere of radius $1$ centered at the origin and another with the same radius centered at $(5,6,7)$. Can one think about them as a single surface?
Some colleagues asked me this question a couple days ago and my answer was no, but then one of them mentioned the example of the two sheet hyperboloid (at one point one of them even said: "so is the set of all surfaces a surface as well?") and I was confused. I looked up the definition of $2$-dimensional manifolds again to make sure, but the answer is still not clear to me.
Sure, there's no need for a manifold to be connected. Your surface might naturally occur as a preimage of a regular value of a mapping $\mathbb R^3\to \mathbb R$, or as the boundary of a $3$-dimensional connected manifold-with-boundary in $\mathbb R^4$, or as the transversal intersection of two connected $3$-dimensional manifolds in $\mathbb R^4$. It's also homeomorphic to the product space $S^2\times S^0$, where $S^2$ is the familiar $2$-sphere and $S^0$ is the $0$-dimensional sphere consisting of two points. So if you wanted to change the definition of "surface" and "manifold" to rule out your surface, then you'd have to change all that surrounding theory, too.