Proof that path-connected spaces are connected

Proof

We argue by contradiction, and restrict a splitting of $X$ to the image of a path.

1. Suppose that $X$ is not connected. Then there are open $U, V \subseteq X$, both nonempty and disjoint, with $U \cup V = X$.
2. Choose $x \in U$ and $y \in V$. Since $X$ is path-connected, there is a continuous $\gamma \colon [0, 1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$.
3. The set $[0, 1]$ is an interval, so it is a connected subset of $\mathbb{R}$, by connected subsets of the real line are intervals. So $[0, 1]$ with its subspace topology is a connected space.
4. By continuous image of a connected set is connected, applied to $\gamma$ on the space $[0, 1]$, the set $K = \gamma([0, 1])$ is connected.
5. The sets $K \cap U$ and $K \cap V$ are open in $K$. They are disjoint, because $U \cap V = \varnothing$, and their union is $K$, because $U \cup V = X$.
6. They are nonempty, because $x = \gamma(0) \in K \cap U$ and $y = \gamma(1) \in K \cap V$.
7. So $K$ is not connected, which contradicts step 4. Thus $X$ is connected.