Compactness of infinite union under these conditions

Question

Assume I have an infinite sequence $(S_k)_{k\in\mathbb N}$ of sets $S_k\subset \mathbb R^n$, assume that all the $S_k$ are compact with respect to the topology induced by some metric $d:\mathbb R^n\times\mathbb R^n\to\mathbb R$.
The infinite union $\bigcup_{k\in\mathbb N} S_k$ is not necessarely compact. But what could one conclude if all the sets $S_k$ have a common point (i.e. $\bigcap_{k\in\mathbb N} S_k \neq \emptyset$) and they are all subset of a given bounded set $\bar S$?

What if all the $S_k$ are distributed around a given point? (for instance the origin)

Answer 1

The conclusion I want to jump to is that the infinite union is compact. But, that is not true.

Suppose $S_n = \{1,\frac1n\}$

and $\bar S = \{\frac1n: \forall n \in \mathbb N\} \cup \{0\}$

$\bar S$ is compact.

$\bigcap S_k = \{1\}$ i.e. non-empty.

There exists a sequence $\in\bigcup S_k$ that converges to $0.$

$0 \notin \bigcup S_k$

$\bigcup S_k$ is not compact.

Answer 2

Hint: $S_k=[-k,k]$

[edit after the question was changed]

If you want $S_k\subset\tilde{S}$ for some compact $\tilde{S}$ then let $S_k$ be the unit line segment intersecting the origin of the plane at angle $\theta=2\pi/k$.