Real Number Line with Point Removed is Not Path-Connected

Theorem
Let $\R$ be the real number line considered as an Euclidean space.

Let $x \in \R$ be a real number.

Then $\R \setminus \left\{{x}\right\}$, where $\setminus$ denotes set difference, is not path-connected.

Proof
We have that $x - 1$ and $x + 1$ are both real numbers, so:
 * $x - 1 \in \R \setminus \left\{{x}\right\}$
 * $x + 1 \in \R \setminus \left\{{x}\right\}$

Let $\mathbb I := \left[{0 \,.\,.\, 1}\right]$ be the closed unit interval.

Suppose there exists a path $f: \mathbb I \to \R \setminus \left\{{x}\right\}$ from $x - 1$ to $x + 1$.

Then by Image of Interval by Continuous Function is Interval, it follows that $x = f \left({s}\right)$ for some $s \in \mathbb I$.

But $x \notin \R \setminus \left\{{x}\right\}$ by definition of set difference.

Hence such an $f$ can not exist.

The result follows by definition of path-connected.