Proving a circumference is not homeomorphic to two adjacent ones

Let $$X$$ be the unit circle in $$\mathbb{R}^2$$; that is $$X=\{(x,y):x^2+y^2=1\}$$ and has the subspace topology. Consider $$Y$$ to be the subspace of $$\mathbb{R}^2$$ given by $$Y=\{(x,y):x^2+y^2=1\}\cup\{(x,y):(x-2)^2+y^2=1\}$$.

Is $$Y$$ homeomorphic to the space $$X$$?

I know I could use path-connectedness to prove $$X$$ and $$Y$$ are not homeomorphic but I am not supposed to. I thought of cardinality, but I really have no clue on how to approach this problem.

Question:

How should I prove $$X$$ and $$Y$$ are not homeomorphic?

I don't know if this is what you have in mind whan you claim that you “could use path-connectedness to prove $$X$$ and $$Y$$ are not homeomorphic”; if it is, I will delete it.
If you remove a point from $$X$$, what remains is connected. But if you remove $$(1,0)$$ from $$Y$$, what remains becomes disconnected.
• @Pedro, but in this case it's easier to show that $Y$ minus the point is not connected directly from the definition of connectedness than using path connectedness. – Ennar Oct 31 '18 at 18:24
• That's easy: the left half of $Y$ is clopen. – José Carlos Santos Oct 31 '18 at 18:28
• The left side of $Y$ is closed becaue it is the intersection of $Y$ with the closed halfplane $\{(x,y)\in\mathbb{R}^2\,|\,x\leqslant0\}$ and it is open because it is the intersection of $Y$ with the open halfplane $\{(x,y)\in\mathbb{R}^2\,|\,x<0\}$. – José Carlos Santos Oct 31 '18 at 18:32