Non-existence of a homeomorphism between two sub-spaces of $\mathbb{R}^{2}$

Question

Let $A_n = \{(x,y)\in\mathbb{R}^{2}:x^2+y^2=\frac{1}{n}\}$ , $n\geq 1$ and let $B = \{(x,0)\in\mathbb{R}^{2}:-1\leq x\leq 1\}$. Prove that: $$ \bigcup_{n=1}^{3}A_n\cup B \mbox{ is not homeomorphic to } \bigcup_{n=1}^{5}A_n\cup B. $$ I tried to suppose the existence of such a homeomorphism and restricted the domain by removing the intersection points between $B$ and the $A_n$ , then I also restricted the codomain by removing the images of the points I've eliminated from the domain. Then this map should also be a homeomorphism, but the number of connected components of the restricted domain is different from the connected components of the restricted codomain. The problem is that it's kind of obvious that the number of connected components is different, but how to prove it? Is there a quicker method to prove the non-existence of such a function? Thank you in advance for every hints or advices (or even better the solution to this problem).