Denote a ``limiting path'' by $ p := \bigcap_{n \in \mathbb{N}} \overline{\bigcup_{\ell \geq n} p_\ell}.$ Show that $p \in P$. Fix $x \neq y \in \Bbb R^n$. Denote by the class of "paths" connecting $x$ and $y$ the collection of sets
$$ P\equiv P(x,y) := \{ p \subset \Bbb R^n: \text{$p$ connected and compact},\ x,y \in p\}.$$
Assume that $(p_n)_{n \in \mathbb{N}} \subset P(x,y)$ is "uniformly bounded" in the sense that there exist $R > 0$ such that $p_n \subset B(0,R)$ for all $n \in \mathbb{N}$.
Denote a "limiting path" by
$$ p := \bigcap_{n \in \mathbb{N}} \overline{\bigcup_{\ell \geq n} p_\ell}.$$
Show that $p \in P$.
My try:
Each $\overline{\bigcup_{\ell \geq n} p_\ell}$ is closed and bounded and connected. [As they are subset of $\overline{B(0,R)}$ and $p_l$ has two points $x,y$ in common.] Then
$$ p := \bigcap_{n \in \mathbb{N}} \overline{\bigcup_{\ell \geq n} p_\ell}.$$ is compact.
I am having trouble to understand why $p$ is connected. In generally for nested connected set, the intersection is not connected. So I am guessing that I have to use compact somehow. Please help.

HINT: If $p$ is not connected, we can write it as $H\cup K$, where $H$ and $K$ are disjoint, non-empty closed sets. There are disjoint open sets $U$ and $V$ such that $H\subseteq U$ and $K\subseteq V$. Let $n\in\Bbb N$; the connectedness of $p_n$ implies that $p_n\nsubseteq U\cup V$ (why?), so there is an $x_n\in p_n\setminus(U\cup V)$. The sequence $\langle x_n:n\in\Bbb N\rangle$ is bounded, so it has a convergent subsequence. Consider the limit of that subsequence.
