Theorem. It is impossible to make ${\Bbb R}^{3} = \underset{\lambda \in \Lambda}{\bigcup} {(S_1)}_{\lambda}$, where each $(S_1)_{\lambda}$ never meets other $(S_1)_{\lambda'}$. That is any two circles $(S_1)_{\lambda} \cap (S_1)_{\lambda'} = \phi$, i.e. an empty set.
Proof. Suppose we have ${\Bbb R}^{3} = \underset{\lambda \in \Lambda}{\bigcup} {(S_1)}_{\lambda}$. Then we can make the surjective morphism $f \colon {\Bbb R}^{3} \to \Lambda$ such that a point $x \in {\Bbb R}^{3}$ maps $\lambda$ iff $x \in (S_1)_{\lambda}$. This is obviously surjective and well-defined, for each circle is disjoint from other circles. I claim that the quotient topology can be endowed on $\Lambda$ by means of $f$, and that $\Lambda$ makes a manifold. Consequently, we have ${\Bbb R}^{3} = S_1 \times \Lambda$ as a manifold. However, we cannot have $\pi_1({\Bbb R}^{3}) = \pi_1(S_1 \times \Lambda) = \pi_1(S_1) \times \pi_1(\Lambda)$, for $\pi({\Bbb R}^{3}) = 0$ whereas $\pi(S_1) = {\Bbb Z}$, hence $0 = {\Bbb Z} \times \pi_1(\Lambda)$, which is a contradiction. Q.E.D.
Is this proof correct?