A specific example of transfinite induction I am trying to understand better how transfinite induction can be applied in different concrete problems. Here is an example that seems relevant, but I am stuck on it. Consider a couple of points $p, q \in \mathbb{R}^2$. I am trying to prove the rather obvious statement that there is an "increasing" neighborhood chain $U_\alpha$ containing $c$ many neighborhoods, ordered by strict inclusion, such that $q \in U_\alpha$, and $p \notin U_\alpha$ for all $\alpha$. Here $c$ is the cardinality of $\mathbb{R}^2$.
I have started reading ordinals and transfinite inductions, so a little help would be highly appreciated. Thanks!
Edit: As pointed out by Porton, the statement I am trying to prove does not require the transfinite induction. I wish to explicitly mention that I know that this statement is rather trivial, I am just trying to understand more about about the principle of transfinite induction and how it applies in different situations.
Further edit: As pointed out by Brian M. Scott, I was wrong in requiring a well-ordered chain. So my question now is, could I construct just a chain of $c$ many elements using transfinite induction? My main curiousity is how to make the jump at the limit ordinal using as little of the topological properties of $\mathbb{R}^2$.
 A: You appear to be trying to construct the family so that it’s not just linearly ordered by strict inclusion, but actually well-ordered: this is impossible, since $\Bbb R^2$ is second countable. To see this, let $\mathscr{B}$ be a countable base for $\Bbb R^2$, and suppose that you have open sets $U_\alpha$ for $\alpha<2^\omega$ such that $U_\alpha\subsetneqq U_\beta$ whenever $\alpha<\beta<2^\omega$. For each $\alpha<2^\omega$ there is an $x_\alpha\in U_{\alpha+1}\setminus U_\alpha$, and there is a $B_\alpha\in\mathscr{B}$ such that $x_\alpha\in B_\alpha\subseteq U_{\alpha+1}$. $\mathscr{B}$ is countable, so there are $\alpha<\beta<2^\omega$ such that $B_\alpha=B_\beta$. But then $B_\alpha\subseteq U_{\alpha+1}\subseteq U_\beta$, so $x_\beta\notin B_\alpha=B_\beta$, which is impossible.
You can get a linearly ordered family of cardinality $2^\omega$, but it won’t be well-ordered, and you don’t need transfinite recursion to construct it. Let $d=\|p-q\|$ be the distance from $p$ to $q$ in the usual metric; then $\{B(p,\epsilon):0<\epsilon<d\}$ is a family of $2^\omega$ open nbhds of $p$, none of them contains $q$, and they are linearly ordered by strict inclusion.
A: No need for transfinite induction to prove this statement.
Just consider the set of (open or closed dependently on your mood) discs with center $p$ and radiuses $]0; |q-p|[$.
