Let $\mathbb{R}_K$ denote the real line in the K-topology, which is the topology generated by the basis $\left\{(a,b)|a,b \in \mathbb{R}\right\} \cup \left\{(a,b)-K|a,b \in \mathbb{R}\right\}$, where $K=\left\{1/n|n \in \mathbb{Z}_+\right\}$. This topology is not path-connected. Suppose there exists a path from $0$ to $1$, which is a continuous function $f:[a,b] \to \mathbb{R}_K$ such that $f(a)=0$ and $f(b)=1$. Because $f$ is continuous, its image is compact and contains $[0,1]$. Since $[0,1]$ is a closed subset of the compact image set, it must be compact, which is a contradiction as there is an open cover $\left\{(-2,2)-K \right\} \cup \left\{(1/n,1)|n \in \mathbb{Z}_+ \right\}$ that does not have a finite subcover.
Can this be shown without using compactness? In other words, is it possible to obtain a contradiction from just the continuity of $f$, $K$ and $K \cup \{0\}$ being closed in $\mathbb{R}_K$, and $[0,1]$ being inside the image of $f$?