I have the following excercise: Let $A$ be an open set. If $x,y\in A$ we write $x\sim y $ when there is a path from $x$ to $y$, this is, $\exists P=\bigcup_{i=0}^{n} [r_{i-1},r_i]$ with $a=r_0$ and $b=r_n$ such that $[r_{i-1},r_i]\subset A$ for $i=0,1,\dots,n$.
Now let $a\in A$, we define $[a]:=\{x\in A: x\sim a\}$. Prove that $[a]$ is connected and open.
To prove that $[a]$ is conneted I consider $x,y\in A$ and since $x\sim a$ and also $y\sim a \implies a\sim y$ since $\sim$ is an equivalence relation (which I already proved), by the latter relation also follows $x\sim z \implies [a]$ is a path connected set.
But how can I prove that $[a]$ is an open set?. My (probably wrong) conjecture is that could be $[a]=A$ and since $A$ is open $[a]$ will be open as well, and that may be the point of the excercise, maybe prove that open sets are connected by paths?; which arise the following question: my textbook says that if $A $ is a subset of $\mathbb{R}^n$ then $A$ is connected, but is this true for any kind of topology?. If it is, then since $x\sim a\; \forall x\in A$ and I would have $A\subseteq [a]$ and the latter inclusion with $[a]\subseteq A$ would imply $[a]=A$; if not how could I prove that $[a]$ is open for any topology?.
My attempt for the last question didn't go too far, because trying to prove that $[a]$ is an open set I would like to show that $\exists \delta >0$ such that $B_{\delta}(x)\subset [a]$. $A$ is open and then there is $\delta$ such that $B_{\delta}(x) \subset A$, does this means that the proof depends of the connectedness of the open ball?. Also this doesn't seems to work taking just one open ball around $x$, but maybe a collection of open balls centered in points of the path given by $x\sim y$ taking care of $\delta$ to have every open ball inside $A$; then if I take $\delta<\frac{1}{2}\operatorname{mín}\{d(x,y):x,y\in \partial(A)\}$ I could cover the path from $x$ to $a$ with open balls each one of them connected and their union will also be an open set included in $[a]$, right?; still taking care of the centers of that balls could be a bit of a hassle, any ideas?.
In this notation $[a]$ is the connected component of $A$ containing $a.$ It is open, because if $x$ is connected to $a,$ take a small disk $D$ around $x$ contained in $A$ (possible because $A$ is open). Now if $y \in D,$ then the (radial) path $\rho$ from $y$ to $x$ is contained in $A$ (since it is contained in $D$) and the path obtained by concatenating the path from $a$ to $x$ with $\rho$ is a path connecting $a$ to $y,$ and so $y\in [a].$