HINT: Suppose that $\langle p_n:n\in\Bbb N\rangle$ is a sequence in $A$ that converges in $\Bbb R^2$. Then there are sequences $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ in $\Bbb R$ such that $p_n=\langle x_n,y_n\rangle$ for each $n\in\Bbb N$. Let $p=\langle x,y\rangle\in\Bbb R^2$ be the limit of the sequence $\langle p_n:n\in\Bbb N\rangle$; you want to show that $p\in A$.
For each $n\in\Bbb N$, $p_n=\langle x_n,y_n\rangle\in A$, so we know that $x_n>0$ and $y_n=\dfrac1{x_n}$, so that $p_n=\left\langle x_n,\frac1{x_n}\right\rangle\;.$
Use the fact that $\langle p_n:n\in\Bbb N\rangle\to\langle x,y\rangle$ to show that $\langle x_n:n\in\Bbb N\rangle\to x$ and $\langle y_n:n\in\Bbb N\rangle\to y$.
Use the continuity of the function $f(x)=\frac1x$ to conclude that $y=\frac1x$ and hence that $\langle x,y\rangle\in A$.
An alternative approach is to show that $\Bbb R^2\setminus A$ is open: show that if $p=\langle x,y\rangle\notin A$, then there is an $r>0$ such that $B(p,\epsilon)\cap A=\varnothing$, where $B(p,\epsilon)$ is the open ball of radius $\epsilon$ centred at $p$. Note that any $\epsilon$ less than or equal to the distance from $p$ to $A$ will work.