How can I show that that compact subsets of $\mathbb{R}$ in this topology are finite subsets. Let $\tau$ be the topology on $\mathbb{R}$ which has as base the collection of all sets of the form 
 $O \setminus C$ where $O \subset \mathbb{R}$ is an open set in the standard topolgy of $\mathbb{R}$ and 
 $C \subset \mathbb{R}$ is a countable subset.
a.Show that $(\mathbb{R},\tau)$ is a connected space which is not locally connected.
b.Find the path component of  $ 0 \in \mathbb{R}$.
c.Show that compact subsets of $\mathbb{R}$ in this topology are finite subsets.
 A: For (c) directly, without (a) and (b):
If $X\subseteq \mathbb R$ is infinite, enumerate some countable sequence of elements of $X$: $x_1,x_2,\dots,x_n,\dots$ and define for $d$ a positive integer: $O_d=\mathbb R\setminus \{x_n: d\mid n\}$.  Then $X\subseteq \bigcup_{d} O_d$. But For any finite set $d_1,d_2,\dots,d_k$, $x_{D}\notin \bigcup_{i=1}^k O_{d_i}$, where $D=d_1d_2\cdots d_k$. So there is no finite subcover.
A: I have tried so far on this problem; but I'm not sure my solution.
a. 
 Suppose $(\mathbb{R}, \tau )$ is disconnected. So there exist $U,V$ such that 
 $U \cup V = \mathbb{R}$, $U \cap V = \emptyset$, $U \neq \emptyset$ ,$U \neq \mathbb{R}$.
 Since $U,V$ open in $\tau$ $U=O_1 \setminus C_1$, $V=O_2 \setminus C_2$ 
$O_1,O_2$ are open, $C_1,C_2$ are countable in standard topology.
$U\cap V = (O_1 \setminus C_1)\cap (O_2 \setminus C_2)= \emptyset$
$\qquad = (O_1\cap O_2)\cap (C_1\cup C_2)^c$
$(O_1\cap O_2)$ is open and $(C_1\cup C_2)$ is countable.
 Let we say $ O=(O_1\cap O_2)$ and $C=(C_1\cup C_2)$. Then we obtain,
$O \setminus C = \emptyset$ 
$O=C$. it's a contradiction. So $(\mathbb{R}, \tau )$ is connected space.
$\mathbb{R} \setminus \mathbb{Q}$ is neighbourhood of $\sqrt{2}$ but  $\mathbb{R} \setminus \mathbb{Q}$
 is not connected. And we can not find connected neighbourhood $V$ of $\sqrt{2}$  such that 
 $\sqrt{2} \in V \subset \mathbb{R} \setminus \mathbb{Q} $. So R is not locally connected.
b. 
 Firstly every open $(a,b)$ is open set in $(\mathbb{R}, \tau )$, C is countable subset. 
 İf $(a,b)\cap C = \emptyset$ then $(a,b)\setminus C =(a,b)$ .
For arbitrary $k \in \mathbb{R}$,there exist a continuous
$$f: [0,1]\rightarrow (\mathbb{R}, \tau)$$ 
 $$x\rightarrow kx$$ 
$f(0)=0$,$f(1)=k $ then $0\sim k$ , $k \in \mathbb{R}$
Then path component of $0$ is $\mathbb{R}$.
c. Let we prove $(\mathbb{R}, \tau )$ is Hausdorff.
 Since every open $(a,b)$ is open set in $(\mathbb{R}, \tau )$ and
  $\mathbb{R}$ is Hausdorff with standard topology. We are done. 
  Let $A\subset \mathbb{R}$ is compact. Then $A$ is closed.
  So I will prove closed subsets of $\mathbb{R}$ is finite. But how?  
