Proving that $\mathbb{R}_K$ is connected with the topology $\tau$ Let $\mathbb{R}_K$ be the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the topology generated by the family of open sets $$\{(a,b),  \, (a,b) \setminus K \mid a \leq b \in \mathbb{R} \}$$ where $$K = \{\frac{1}{n} \mid n \in \mathbb{N} \}$$
Prove that the usual topology of $\mathbb{R}$ is strictly contained in  $\tau$. 
Prove that $\mathbb{R}_K$ is connected but not arc connected. 
$\textbf{My attempt}$
Clearly the euclidean topology is contained strictly in $\tau$ because one can consider the open set $(-1,1) \setminus K$ in the topology $\tau$ as a neighborhood of $0$, and there is no usual open set $A$ such that $0 \in A$ and $A \subset (-1,1) \setminus K$
It's not arc connected , in fact consider any continuous map $\gamma : [0,1] = I \to R_K$ joining $0$ and $1$. $A := \gamma^{-1}((-1,1) \setminus K)$ must be open in $I$. So there must be an interval like $[0,\varepsilon) \subset A$ with $\gamma(\varepsilon) >0$. But this can't be true, since if it was, we would have that $\gamma([0,\varepsilon)) \subset (-1,1) \setminus K $. Now since $\gamma$ is also continuous in the usual topology, intervals are mapped into intervals, so all the points $\frac{1}{n}$ for $n$ big should be in $\gamma([0,\epsilon))$ and this is not possible.
Now I can't prove that it has to be connected.
Thanks in advance for anyone who wants to give me some help!
 A: $A = \{x \in \Bbb R: x>0\}$ is connected as it has the same topology as a subspace of $\Bbb R$ and of $\Bbb R_K$, and the same holds for $B=\{x \in \Bbb R: x < 0\}$.
Now, taking the closures in $\Bbb R_K$, it's easy to see that $\overline{A}=[0,+\infty)$ and $\overline{B}=(-\infty,0]$ and as closures of connected sets, they are connected too in $\Bbb R_K$. They intersect in $0$ so their union, i.e. $\Bbb R_K$ is connected as well.
A: Suppose that $A$ and $B$ partition $\mathbb{R}$ and are clopen disjoint. Suppose that there is an integer $n$ such that the interval $(1/(n+1),1/n)$ contains points from both $A$ and $B$: this is a contradiction, since the intervals of that form are clearly connected (the induced topology on them is the standard one). Now suppose that there is $n$ such that $(1/(n+2),1/(n+1))\subset A$ and $(1/(n+1),1/n)\subset B$. Again, this is a contradiction, since the point $1/(n+1)$ could not belong to either $A$ or $B$ (everyone of its basic open neighbourhoods intersects both $A$ and $B$). So all of the intervals $(1/(n+1),1/n)$ belong to, say, A, and so all of the points $1/n$. But then, $0$ belongs to $A$ as well, since every open neighborhoods intersects $A$. 
This means that the whole interval $[0,1)$ is in $A$, so we can restrict our attention to $\mathbb{R}\setminus [0,1)$. But since on this set the topology is the usual one, we have a contradiction.
A: Hint: $(0,+\infty)$ and $(-\infty,0)$ are connected in $\mathbb{R}_K$. Consider the connected component which contains $0$.
