Why this: relative topology of F is discrete for being F a countable subset, which contradicts its compacity (of F)?

Question

Let's see $\Bbb R$ with the countable complement topology is not path connected. Let's suppose there is a function $f:[0, 1] \to \Bbb R$ which connects two points $x$ and $y$. If it was a path, then $K := f([0, 1])$ would be compact and connected. If $K$ was not finite, it would have a countable subset $F$, and then closed, which implies compact for being a closed in a compact. However, relative topology of $F$ is discrete for being $F$ a countable subset, which contradicts its compacity. Can someone explain me the last phrase (However,...)? I understand the case of finite.

Answer 1

You have a countable $F\subseteq\Bbb R$ in the co-countable topology. Let $x\in F$, and let $V_x=\{x\}\cup(\Bbb R\setminus F)$. Then $\Bbb R\setminus V_x=F\setminus\{x\}$, which is countable, so $V_x$ is open. And $V_x\cap F=\{x\}$, so $\{x\}$ is a relatively open subset of $F$. This is true for every $x\in F$, so the relative topology on $F$ is discrete. In particular, if $F$ is infinite, $\{V_x:x\in F\}$ is an open cover of $F$ with no finite subcover, and $F$ therefore cannot be compact.