(a) The topology is strictly finer that the standard topology on $[0,1]$ which is compact and Hausdorff, therefore, it is not compact. Cool, huh? Another way to show this directly is as follows: $[0,1]=([0,1]-K)\cup\cup_i(1/(i+1),1]$ has no finite subcovering. Yet, another way is just to say that $K$ is a closed subspace of $[0,1]$ in $\mathbb{R}_K$ which is not compact (the set is infinite and all points are isolated), therefore, $[0,1]$ is not compact.(b) Both topologies induce the same topology on $(-\infty,0)$ and $(0,+\infty)$ (it is clear for the first one, while for the second a subset $U$ is open in the subtopology of $\mathbb{R}_K$ iff it equals the intersection of the subspace with an open set $V$ in $\mathbb{R}$ minus some points in $K$ iff it equals the intersection of the subspace with $V-\{0\}$ minus those points in $K$ iff it is open in the subtopology of $\mathbb{R}$ ), therefore, both are connected in either topology. Their closures are connected and share the common point $0$ , therefore, their union — the whole space — is connected.(c) Suppose there is a path from 0 to 1 in $\mathbb{R}_K$ . It is a continuous function from a compact connected space, therefore, the image is compact and connected in $\mathbb{R}_K$ . Hence, it must also be connected as a subspace of $\mathbb{R}$ (as it has a coarser topology), i.e. the image is convex and contains the whole interval $[0,1]$ . This implies that $[0,1]$ is a closed subspace of the compact image, i.e. it is compact in $\mathbb{R}_K$ . Contradiction.