$ \mathbb{R} $ with the $ K $-topology ($ \mathbb{R}_{K} $) Recall that $ \mathbb{R}_{K} $ denotes $ \mathbb{R} $ with the $ K $-topology.
(a) Show that $ [0,1] $ is not compact as a subspace of $ \mathbb {R}_{K} $
(b) Show that $ \mathbb{R}_{K} $ is connected. [Hint: $ (- \infty, 0) $ and $ (0, \infty) $ inherit their usual topologies as subspaces of $ \mathbb{R}_{K} $.]
(c) Show that $ \mathbb{R}_{K} $ is not path connected.
For part (a) I found the following test, but I couldn't understand it well.
"Recall that the K-topology is generated by the basis {$(a, b), (a, b) − K | a < b$}, where
K = { $\frac{1}{n}| n\in Z^+$}  for each i ,let $U_i = (\frac{1}{i} ,2) U (-1,1) - k:$ the open cover {${u_i}$} of $[0,1]$ does not have finite subcover so$ [0,1]$  is not compacts"
Is this correct? On the other hand, I have not been able to conclude anything about (b) and (c). Any help please.
 A: The cover you referred to is (improved version) $\{(-2,2)-K\} \cup (\frac{1}{n},1)_{n \in N^+}$ which is an open cover of $[0,1]$ ($0$ and $1$ are covered by the first member, any $x>0, x < 1$ will be covered by some $(\frac{1}{n}, 1)$; and it's clear that all these sets are open in $\Bbb R_K$ by its definition. 
Now if we have a finite subcover of it, we only use finitely many sets of the second type, so there is one with the largest $n$, say $\frac{1}{N}$ is that point. But then it's easy to find a point (or lots of points actually) that cannot be covered by this finite subcover, e.g. $\frac{1}{N+1}$ is one. (it's not covered by the first set, nor by any of the finitely many others by the choice of $N$.)
In $\Bbb R_K$ the sets $L = ( -\infty, 0)$ and $R = (0,+\infty)$ have the same topology as subspaces as they had in $\Bbb R$ in the usual topology. In particular, they are connected in $\Bbb R_K$ and both sets only add $0$  to their closure. So $\Bbb R_K  = \overline{L} \cup \overline{R}$ is a union of two intersecting connected sets and thus connected. 
Suppose there were a continuous path $p: [0,1] \to \Bbb R_K$ such that $p(0)=0, p(1)=1$. Then by connectedness of the image, $p[[0,1] \supseteq [0,1]$ and so $[0,1]$ is a closed subset of the compact image set, hence compact. But this contradicts what we’ve seen before that it is not compact. So $\Bbb R_K$ is not path-connected. 
