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The question says: Let $Y$ be the subspace of $\mathbb{R^2}$ given by \begin{equation} Y=\{(x,y):x^2+y^2=1\}\cup\{(x,y):(x-2)^2+y^2=1\} \end{equation} Is $Y$ homeomorphic to $S^1$? And to an interval? My reasoning is that the first circle in $Y$ namely $x^2+y^2=1$ gets mapped to all the points in $S^1$. Now the remaining points of the circle with equation $(x-2)^2+y^2=1$ must me mapped somewhere in $S^1$. So this means that two different points from $x^2+y^2=1$ and $(x-2)^2+y^2=1$ are mapped to the same point so there can't be a bijection from $Y$ to $S^1$. Is there something wrong with my reasoning? For the second part I don't know where to start. Maybe showing that some transitive property doesn't apply here but I am not really sure.

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    $S^1$ minus any point is connected. Is the same true for $Y$? – 2017-02-13

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$Y$ is the union of two circles $C,C'$ that are tangent to each other at the point $a = (1,0)$. If you remove the point $a$ from $Y$, then the result is a topological space $\tilde{Y}$ with two path components. But removing a single point from the unit circle $S$ would result in a topological space $\tilde{S}$ with just one path component. If $Y$ were homeomorphic to $S$, then $\tilde{Y}$ would have to be homeomorphic to $\tilde{S}$, which is impossible.

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It is true that the two spaces are not homeomorphic; however, your reasoning is incorrect (or rather, not fully justified). Specifically, when you write

My reasoning is that the first circle in $Y$ namely $x^2+y^2=1$ gets mapped to all the points in $S^1$,

how do you know this? In general, you can have spaces $A$ and $B$, and a proper subspace $C\subsetneq A$, such that $C\cong B$ but also $A\cong B$; just because you can map $C$ onto all of $B$, doesn't mean you have to. That is, in principle we could have a homeomorphism that only mapped $Y$ onto some of $S^1$; just because the obvious map takes $Y$ onto all of $S^1$, doesn't mean every map has to.

Now of course, it is true that every injective continuous map from $S^1$ to $S^1$ is onto, so the claim you make is correct; but this takes proof.

A simpler approach would be to find some fact about one space, which is not true of the other. Connectedness is usually a good place to start. Now, both these spaces are connected; but what happens when you remove a point from either space?