The intersection with every line L in $\mathbb{R^2}$ is open in $L$ with the topology
$\mathcal{T}_L = \{G\cap L : G \text{ open in } \mathbb{R^2} \text{ with the Euclidean Topology } \} $, but the set is not open in $\mathbb{R^2}$ with the Euclidean Topology ?
I couldnt find and such set but am pretty sure that there exist such a set , if you have any ideas let me know ! thanks in advance !
On
By a classical result of Mazurkiewicz cited in this paper by Mauldin, assuming the axiom of choice there is a so-called "$2$-point set", i.e., a set $S\subset\mathbb R^2$ which intersects every line in exactly two points. In particular a $2$-point set intersects every line in a closed set, so its complement has your property, it intersects every line in a relatively open set.
By a result of Larman quoted in this MathOverflow question, a $2$-point set can't be an $F_\sigma$ set,
so its complement can't be a $G_\delta$ set, much less an open set. It can easily be arranged that the
$2$-point set is not even a Borel set, and in fact it's a well known open problem whether a $2$-point set can be a Borel set.
(To kick it from the unanswered queue)
Summarizing the comments:
Take $A=\Bbb R\setminus P$ where $P$ is a parabola with one point removed.