Suppose $X$ is path-connected. We prove $X$ is connected by contradiction.

Assume $X$ is disconnected. Then there exist nonempty disjoint open sets $U, V$ with $X = U \cup V$.

Since $U$ and $V$ are nonempty, choose $a \in U$ and $b \in V$.

Since $X$ is path-connected, there exists a continuous map $\gamma: [0, 1] \to X$ with $\gamma(0) = a$ and $\gamma(1) = b$.

Now consider the two preimage sets: $A = \gamma^{-1}(U), \qquad B = \gamma^{-1}(V).$

$A$ and $B$ are open in $[0, 1]$: Since $\gamma$ is continuous and $U, V$ are open in $X$, the preimages $A = \gamma^{-1}(U)$ and $B = \gamma^{-1}(V)$ are open in $[0, 1]$.

$A$ and $B$ are nonempty: We have $0 \in A$ (since $\gamma(0) = a \in U$) and $1 \in B$ (since $\gamma(1) = b \in V$).

$A$ and $B$ are disjoint: If $t \in A \cap B$, then $\gamma(t) \in U \cap V = \emptyset$, which is impossible.

$A$ and $B$ cover $[0, 1]$: For any $t \in [0, 1]$, $\gamma(t) \in X = U \cup V$, so $t \in A \cup B$.

Therefore $\{A, B\}$ is a separation of $[0, 1]$.

But $[0, 1]$ is connected (it is an interval in $\mathbb{R}$, and by Theorem 4.13, intervals are connected). This is a contradiction.

Therefore $X$ must be connected.

Fix $p \in X$ and let $U = \{x \in X : \text{there is a path from } p \text{ to } x\}$.

$U$ is open: Let $x \in U$. Since $X$ is locally path-connected, there is a path-connected open neighborhood $V$ of $x$. For any $y \in V$, concatenate a path from $p$ to $x$ with a path from $x$ to $y$ in $V$. Thus $V \subseteq U$, and $U$ is open.

$X \setminus U$ is open: Let $x \in X \setminus U$. There is a path-connected open neighborhood $V$ of $x$. If any $y \in V$ belonged to $U$, then concatenating a path from $p$ to $y$ with a path from $y$ to $x$ in $V$ would give a path from $p$ to $x$, contradicting $x \notin U$. So $V \subseteq X \setminus U$, and $X \setminus U$ is open.

Since $X$ is connected and $U$ is nonempty (contains $p$) and clopen, $U = X$.