TITLE: $A\subset \mathbb{R}$ with more than one element and $A/ \{a\}$ is compact for a fixed $a\in A$
QUESTION [5 upvotes]: Question is :
Suppose $A\subset \mathbb{R}$ with more than one element and $A/ \{a\}$ is compact for a fixed $a\in A$ then 

$A$ is compact
Every subset of $A$ must be compact
$A$ must be finite
$A$ is disconnected

Only compact subsets of $\mathbb{R}$ I can think of are finite union of closed intervals and finite sets.
Take a  finite union of closed intervals, If i remove one element for that I would end up with something which contains $[a,b)$  and this would be not compact so should be the whole set.
So, First option is wrong i.e., $A$ is not compact.
I can not say anything about second and fourth option but third option is possible I guess.
Please help me to see this in detail..

REPLY [4 votes]: Let $\{U_i\}_{i\in I}$ be an open cover for $A$. This is also an open cover for $A\setminus\{a\}$. As this is compact, there exists a finite subcover. Together with one $U_i$ with $a\in U_i$ we obtain a finite subcover for $A$. Hence, yes, $A$ is compact.
No. Consider $A=[0,1]\cup\{2\}$, which has the noncompact subset $(0,1)$.
No. Consider $A=[0,1]\cup\{2\}$ again, which is infinite.
Yes. Consider the open cover $$A\setminus\{a\}\subseteq\bigcup_{r>0}(\mathbb R\setminus[a-r,a+r]).$$
By compactness there is a finite subcover, which boils down to $A\setminus\{a\}\subseteq \mathbb R\setminus[a-r,a+r]$ for some $r>0$. Thus $a$ is an isoloated point of $A$ and $A$ is disconnected.