(Getting another question off the unanswered list; I will as usual leave a little for the reader to do.)
For this problem the characterizations of the subspace and quotient topologies given in the question are not especially useful; the ones suggested by J. Damé in the comments are much more helpful, because they give explicit descriptions of the open sets in these topologies.
Let $\tau$ be the usual topology on $\Bbb R^2$, and for convenience let $p=\langle 0,0\rangle$. Let $U\subseteq S$, and suppose first that $p\notin U$.
- Show that $U\in\tau_1$ iff $U\in\tau$ iff $U\in\tau_2$.
Now suppose that $p\in U$.
- Show that $U\in\tau_1$ iff there is a $V\in\tau$ such that $p\in V$ and $V\cap S=U$.
- Show that $U\in\tau_2$ iff there is a $V\in\tau$ such that $\{0\}\times\Bbb R\subseteq V$ and $V\cap S=U$. Show further that $\operatorname{cl}_{\tau_2}U=S\cap\operatorname{cl}_\tau V$; this will be useful in showing that $\tau_2$ is regular.
These observations let us determine exactly what sets are in $\tau_1$ and what sets are in $\tau_2$. In particular it should be clear that $\tau_2\subsetneqq\tau_1$, since every $\tau$-open nbhd of the $y$-axis is a $\tau$-open nbhd of the origin, but there are clearly $\tau$-open nbhds of the origin that do not contain the $y$-axis.
Every metrizable space is regular, so we’d be done if we could show that $\tau_2$ is metrizable. Unfortunately, it is not: it is not even first countable. Specifically, the space $\langle S,\tau_2\rangle$ does not have a countable local base at the point $p$.
To see this, suppose that $\mathscr{B}=\{B_n:n\in\Bbb Z^+\}$ is a family of open nbhds of $p$. For each $n\in\Bbb Z^+$ there is a $k_n\in\Bbb Z^+$ such that $\left\langle\frac1{k_n},n\right\rangle\in B_n$. Let $F=\left\{\left\langle\frac1{k_n},n\right\rangle:n\in\Bbb Z^+\right\}$, $V=\Bbb R^2\setminus F$, and $U=V\cap S$. $F$ is closed in $\left\langle\Bbb R^2,\tau\right\rangle$ and disjoint from the $y$-axis, so $V$ is a $\tau$-open nbhd of the $y$-axis, and $p\in U\in\tau_2$. But $\left\langle\frac1{k_n},n\right\rangle\in B_n\setminus U$ for each $n\in\Bbb Z^+$, so $B_n\nsubseteq U$ for each $n\in\Bbb Z^+$, and $\mathscr{B}$ is therefore not a local base at $p$.
The topology $\tau_1$, on the other hand, is metrizable: it’s easy to check that if $d$ is any metric on $\Bbb R^2$ that generates the usual topology $\tau$, then $d\upharpoonright(S\times S)$ is a metric on $S$ that generates the subspace topology $\tau_1$.
Finally, $\tau_2$ is regular. It suffices to show that if $x\in S$, and $x\in U\in\tau_2$, then there is a $G\in\tau_2$ such that $x\in G\subseteq\operatorname{cl}_{\tau_2}G\subseteq U$. This is easy to see if $x\ne p$, so suppose that $p\in U\in\tau_2$. There is a $V\in\tau$ such that $\{0\}\times\Bbb R\subseteq V$ and $V\cap S=U$. $\left\langle\Bbb R^2,\tau\right\rangle$ is normal, and $\{0\}\times\Bbb R$ is $\tau$-closed, so there is a $W\in\tau$ such that $\{0\}\times\Bbb R\subseteq W\subseteq\operatorname{cl}_\tau W\subseteq V$. Let $G=W\cap S$; then $G\in\tau_2$, and
$$p\in G\subseteq\operatorname{cl}_{\tau_2}G=S\cap\operatorname{cl}_\tau W\subseteq S\cap V=U\,,$$
as desired.
Best Answer
Let $A=\{\,(x,y)\,|\,\, x^2+y^2\leq1 \text{$\,\,\,$and$\,\,\,$} (x,y)\in S\,\}$. Then $A$ is compact with respect to $\tau_2$ but not with respect to $\tau_1$.
My reasoning:
$A$ is not open in usual topology, so it is not compact. Since, $\tau_1$ is usual topology. Furthermore, we can also imagine an open covering which has no sub-covering.
Key idea: "Any open set in $\tau_2$ which contains the origin $(0,0)$, must contain a strip like $$U_{\epsilon}=((-\epsilon,\epsilon)\times\Bbb{R}^1)\cap S$$ due to the definition of the quotiend topology." So, given an open covering of $A$ in $\tau_2$, it contains an open set $U$ which contains such an $U_{\epsilon}$. Then consider the closed and bounded set $B=A-U_{\epsilon}$. It does not contain $(0,0)$, so for it the topology is the usual one. So we can find a finite subcover for it from the given covering. This subcover together with $U$ is a subcovering for $A$.
I need to improve my reasoning. I hope my example is correct.