Find the smallest topology on the unit circle $S^1$ which makes $\alpha$ continuous I need to find the smallest topology on $S^1$ which makes the following function $\alpha:S^1\rightarrow\mathbb{R}:\alpha(x,y)=x$ continuous. This is what I have so far:
$(x,y)=(x,\pm\sqrt{1-x^2})\in S^1$. But because we need the inverse here I'm guessing I just should take the plus sign. For the inverse $\alpha^{-1}:\Bbb R \to S^1$ we have $\alpha^{-1}(x) = (x,\sqrt{1-x^2})$. Now we define the set
$$
S := \{U \subset S^1: \exists U\in\tau: U=\alpha^{-1}(U) \}.
$$
Because $x\in[0,1], S=[0,1]\times[0,1]$. Now the book says that the smallest topology on $S^1$ that makes $\alpha$ continuous is the topology generated by $S$. So
$$
\langle S \rangle := \hskip{-2em} \bigcap_{\tau\text{-topology on $S^1$ containing $S$}}\hskip{-3em} \tau
$$
However I have no idea how to find this. Thinking about $S=[0,1] \times [0,1]$, I was thinking about the Euclidean Topology, but this is not possible because I also have to show that it is not Hausdorff, and any metric space is Hausdorff. Can I get some push in the right direction, because I think i'm overcomplicating the problem.

The topology will be generated by inverse images of open sets in $[-1,1]$. Open intervals in the interior of this interval will have two open arcs as preimage. The inverse image of a neighborhood of an endpoint will be a symmetric open arc around the point $(1,0)$ or $(-1,0)$. In general, the topology that these preimages generate consists of standard open sets in $S^1$ with the additional property that they are symmetric in the $x$-axis. This is not a Hausdorff topology since no open sets can separate a point from its reflection in the $x$ axis.
