If every An is connected, then A is connected I'm practicing for a midterm, so I've been finding exams with answer keys online and solving through them. The last question of this one reads as follows:
Let $R^2$ be given the usual metric topology. Suppose that $(A_n)_n$ is a decreasing
sequence of nonempty, compact subsets of $R^2$ (i.e. $A_n$$_+$$_1$ ⊆ $A_n$). We have shown previously that $A := ∩_nA_n$ is a nonempty, compact set. We have also shown that if every $A_n$ is connected, then $A$ is connected. Suppose that each $A_n$ is path connected. Is $A$ path connected? If so, prove it is; if not, give a counterexample.
I think that the sentence "We have also shown that if every $A_n$ is connected, then $A$ is connected" is referring to something done in class, as I'm not finding this question in the exam itself. I'm not sure I understand why it's true, so how would one prove it?
 A: Suppose $A \subset U$, where $U$ is open. Then there is some $N$ such that
$A_n \subset U$ for $n \ge N$.
To see this, note that $B_n = A_n \cap U^c$ is a nested collection of compact sets
and hence if $B_n$ is non empty for all $n$ then so is $B = \cap_n B_n = A \cap U^c$, which is a contradiction.
Now suppose $U,V$ are open, disjoint and $A \subset U \cup V$. Then we must have
$A_n \subset U \cup V$ for $n$ sufficiently large. Since $A_n$ is connected,
we must have either $A_n \subset U$ or $A_n \subset V$. Hence we must
have either $A \subset U$ or $A \subset V$ and so $A$ is connected.

Counterexample showing that path connectedness of the $A_n$ does not 
imply $A$ is path connected.
Let $A = \{0\} \times [-1,1] \cup \{ (x, \sin {1 \over x} \}_{x  \in (0,1]}$. This is
the 'closed topologist's sine curve' and is compact, connected, but not
path connected.
Let $A_n = A \cup [0,{1 \over n}] \times [-1,1] $. Then $A_n$ is compact, path connected, and
$A = \cap_n A_n$.
The issue is that while each $A_n$ is path connected, the length of the path joining
(for example) $(0,0)$ and $(1, \sin 1)$ is unbounded as $n \to \infty$.
