If $x\in U$ let $A=\{y\in U\}$ such that there is a path between $x$ and $y$ . Using 8(d), $A\subseteq U$ is path connected. If $y\in A$ then there is a basis neighborhood $V\subseteq U$ of $y$ , $V$ is a ball, and $y$ is path connected with any point in $V$ (just take the closed line interval connecting the points). Therefore, $V\subseteq A$ and $A$ is open. For similar reason, if $z\in U-A$ then there is a neighborhood $V$ of $z$ contained in $U$ . If some point in $V$ were path connected to $x$ then so would be $z$ , but $z\notin A$ . Therefore, $U-A$ is open. Then $A$ is closed in $U$ . $A$ is not empty as $x\in A$ , so if $U-A$ is not empty as well, then there is a separation of $U$ . Therefore, $U=A$ is path connected.