= Solution
Every point has arbitrarily small <path-connected> open neighbourhoods. Indeed, in a <coordinate chart> homeomorphic to $\mathbb R^n$, take the inverse image of a sufficiently small open <Euclidean ball>. The ball is <convex>, hence <path-connected>, and its inverse image is <path-connected> by the <homeomorphism>. If the supplied neighbourhood is not itself open, first take an open neighbourhood within it and a ball contained in its chart image. This proves that $X$ is a <locally path-connected space>.
Fix $x_0\in X$ and let $A$ consist of the points joined to $x_0$ by a <path>. For $x\in A$, a <path-connected> open neighbourhood $U_x$ is contained in $A$: concatenate a <path> from $x_0$ to $x$ with one from $x$ to any point of $U_x$. Hence $A$ is <open>. For $x\notin A$, the same kind of neighbourhood cannot meet $A$, since a meeting point would provide a <path> to $x$. Hence $X\setminus A$ is also <open>. Since $X$ is a <connected space> and $A$ is nonempty, $A=X$. This proves \b[$X$ is path-connected], including the general principle that <connected locally path-connected spaces are path-connected>. The empty-space case, if allowed, is vacuous.
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