Lets say I have defined $S^{1}$ as $[0,1]/\sim$ where $x\sim y$ when $x=y$ except when $x,y$ equal 0 or 1. In other words, $[0,1]/\sim$ is the singletons and the set $\{0,1\}$. I want to show that this is a manifold directly. Let $[p]$ denote the equivalence class of $p$ and let $p$ denote the canonical surjection. Taking inspiration from the proof using a stereographic projection here is my attempt:
Let $\phi([t])=\frac{cos(2\pi t)}{1-sin(2\pi t)}$ and let $\psi([t])=\frac{cos(2\pi t)}{1+sin(2\pi t)}$, where $\phi$ is defined on $U_{1}=S^{1}\backslash \{[1/4]\}$ and $\psi$ is defined on $U_{2}=S^{1}\backslash \{[3/4]\}$. These two functions are well clearly well defined. Furthermore, $p^{-1}(U_{1})=[0,1/4)\cup (1/4,1]$ and $p^{-1}(U_{2})=[0,3/4)\cup (3/4,1]$. Both sets are the union of two open sets in $[0,1]$ and hence open. Note $\phi\circ p = \frac{cos(2\pi t)}{1-sin(2\pi t)}$. Similar is true for $\psi\circ p$. These two functions are continuous on $[0,1/4)\cup (1/4,1]$ and $[0,3/4)\cup (3/4,1]$ respectively. Both functions are surjective with $\mathbb{R}$ and are injective as well. As such $\phi$ and $\psi$ are homeomorphic with $\mathbb{R}$. In other words, $S^{1}$ is locally Euclidean.
Now where I am stuck is in showing $S^{1}$ is Hausdorff and second countable directly. I am thinking that it is not that hard to show an explicit homeomorphism between $[0,1]/\sim$ and the graph of the circle in $\mathbb{R}^{2}$. I could then use that the graph is a subset of $\mathbb{R}^{2}$ hence Hausdorff and second countable to show that $[0,1]/\sim$ is Hausdorff and second countable but I was wondering if there is a more direct way.
Edit: I think I found out how to show Hausdorff directly. Let $[x]$ and $[y]$ be two points in $S^{1}$. Assume that neither are $[0]=[1]$. Consider then its inverse under $p$ which is $x,y$ respectively. By the fact that $[0,1]$ is Hausdorff, there are open sets $U,V$ such that $x\in U$ and $y\in V$ where $U$ and $V$ are disjoint. Since $[0,1]$ is a metric space we can ensure that $0,1\notin U,V$. As p is injective on $(0,1)$ we have that $p(U)=U'$ and $p(V)=V'$ and $p^{-1}(U')=U$ and $p^{-1}(V')=V$ meaning $U',V'$ are open. Its not hard to show they are disjoint and $U'$ contains $[x]$ and $V'$ contains $[y]$.
On the otherhand assuming $[x]=[0]=[1]$ we can find open sets $U',U''$ that contain y and that are disjoint from open sets that contain $0$ and $1$. Let $U$ be the intersection of $U'$ and $U''$ which is open and must contain $y$. Let $V$ be the union of the two open sets that contain $0$ and $1$ repsectively. For the same reason as above $p(U)$ is open. V can be wlog assumed to be of the form $[0,\alpha)\cup (\beta,1]$. $p^{-1}p(V)=\{x\in [0,1]| p(x)\in p(V)\}$. If $x \notin V$ but in $p^{-1}p(V)$ that would mean that $p(x)\in p(V)$ but that would imply that there is $y\in V$ such that $p(x)=p(y)$ but $x\neq y$ meaning $x=0$ or $1$ but $0,1\in V$. In otherwords $p^{-1}p(V)$ is open meaning $p(V)$ is open. For the same reason as above $p(U)$ is disjoint from $p(V)$ hence we are done.
I am still stuck on showing second countability however and would appreciate a hint.