First let us prove that $Y$ has no subspaces homeomorphic to $\mathbb{S}^1$. In order to reach a contradiction suppose $A\subseteq Y$ is homeomorphic to $\mathbb{S}^1$. Notice that $Y\setminus\{(0,0)\}$ has three connected components, which are $Y_1=\{(0,y):0, $Y_2=\{(0,y):-1\leq y<0\}$ and $Y_3=\{(x,0):0. This means that $A\setminus(0,0)$, being connected, is contained in exactly one of the sets $Y_i,i\in\{1,2,3\}$. Hence $A$ is contained in exactly one of the sets $Y_i\cup\{(0,0)\},i\in\{1,2,3\}$. Because all of the $Y_i\cup\{(0,0)\}$ are homeomorphic to a (closed) interval and any connected subset of an interval is an interval, $A$ is homeomorphic to an interval containing more than one point. This is a contradiction, because removing any point from $A$ does not disconnect it, but any interval containing more than one point has a point such that the interval minus this point is disconnected (such a point is called a cut point).
Then let us prove that $\mathbb{S}^1$ has no subspaces homeomorphic to $Y$. Again, to reach a contradiction suppose $B\subseteq\mathbb{S}^1$ is homeomorphic to $Y$. Notice that $B\neq \mathbb{S}^1$, since $\mathbb{S}^1$ does not have cut points but $B$ has. This means that there is a point $p\in\mathbb{S}^1$ such that $B\subseteq\mathbb{S}^1\setminus\{p\}$. But $\mathbb{S}^1\setminus\{p\}$ is homeomorphic to an (open) interval, so we can deduce that $B$ is homeomorphic to an interval as before. This is a contradiction because $B$ has a point $q$ (corresponding to $(0,0)\in Y$) such that $B\setminus\{q\}$ has three components (corresponding to $Y_1$, $Y_2$ and $Y_3$), but an interval which has a point removed has at most two components.