[proofplan] We use the neighborhood-basis characterization of local path-connectedness. Given a point $x \in U$ and an open neighborhood $V$ of $x$ inside $U$, the openness of $V$ in the [subspace topology](/page/Subspace%20Topology) and the openness of $U$ in $\mathbb R^n$ allow us to find a Euclidean ball centered at $x$ and contained in $V$. That ball is convex, hence path-connected by the same-run theorem that [convex subsets of Euclidean space are path-connected](/theorems/9157). Therefore every neighborhood of every point contains an open path-connected neighborhood of that point. [/proofplan] [step:Reduce local path-connectedness to finding path-connected open neighborhoods inside arbitrary neighborhoods] We use the following formulation of local path-connectedness: a [topological space](/page/Topological%20Space) is locally path-connected if every point has a neighborhood basis consisting of open path-connected sets. Equivalently, it is enough to prove that for every $x \in U$ and every open subset $V \subset U$ with $x \in V$, there exists an open path-connected subset $W \subset U$ satisfying $x \in W \subset V$. Fix $x \in U$ and fix an open subset $V \subset U$ with $x \in V$. We will construct such a set $W$. [/step] [step:Choose a Euclidean ball around $x$ contained in the given neighborhood] Since $V$ is open in the subspace topology on $U$, there exists an open subset $O \subset \mathbb R^n$ such that $V = U \cap O$. Since $x \in V$, we have $x \in O$. Because $O$ is open in the Euclidean topology on $\mathbb R^n$, there exists $r > 0$ such that the Euclidean open ball \begin{align*} B(x,r) := \{y \in \mathbb R^n : |y - x| < r\} \end{align*} satisfies $B(x,r) \subset O$. Also $x \in U$, and $U$ is open in $\mathbb R^n$. Hence there exists $s > 0$ such that $B(x,s) \subset U$. Define \begin{align*} \rho := \min\{r,s\}. \end{align*} Then $\rho > 0$, and \begin{align*} B(x,\rho) \subset O \cap U = V. \end{align*} Set $W := B(x,\rho)$, regarded as a subset of $U$. Since $B(x,\rho)$ is open in $\mathbb R^n$ and $B(x,\rho) \subset U$, it is open in the subspace topology on $U$. [guided] We need a path-connected [open set](/page/Open%20Set) inside the arbitrary neighborhood $V$. Since $V$ is open only as a subset of $U$, we first translate that statement back into Euclidean openness. By definition of the subspace topology, there is an open subset $O \subset \mathbb R^n$ such that \begin{align*} V = U \cap O. \end{align*} Because $x \in V$, we have $x \in O$. The Euclidean openness of $O$ then gives a radius $r > 0$ with \begin{align*} B(x,r) := \{y \in \mathbb R^n : |y - x| < r\} \subset O. \end{align*} This only places the ball inside $O$, not necessarily inside $U$. We also use the hypothesis that $U$ itself is open in $\mathbb R^n$. Since $x \in U$, there is a radius $s > 0$ such that \begin{align*} B(x,s) \subset U. \end{align*} Taking \begin{align*} \rho := \min\{r,s\} \end{align*} ensures both containments simultaneously. Since $r > 0$ and $s > 0$, we have $\rho > 0$. Therefore \begin{align*} B(x,\rho) \subset B(x,r) \cap B(x,s) \subset O \cap U = V. \end{align*} Now define $W := B(x,\rho)$, viewed as a subset of $U$. Because $B(x,\rho)$ is open in $\mathbb R^n$ and is contained in $U$, it is open in the subspace topology on $U$. Thus $W$ is an open neighborhood of $x$ in $U$ and satisfies $W \subset V$. [/guided] [/step] [step:Show the chosen ball is path-connected] We show that $W = B(x,\rho)$ is convex. Let $y,z \in W$, and let $t \in [0,1]$. Define the point \begin{align*} p_t := (1-t)y + tz \in \mathbb R^n. \end{align*} Using the triangle inequality and the defining inequalities $|y-x| < \rho$ and $|z-x| < \rho$, we get \begin{align*} |p_t - x| = |(1-t)(y-x) + t(z-x)| \le (1-t)|y-x| + t|z-x| < (1-t)\rho + t\rho = \rho. \end{align*} Thus $p_t \in B(x,\rho)$ for every $t \in [0,1]$, so $B(x,\rho)$ is convex. By [citetheorem:9157], every convex subset of $\mathbb R^n$ with the subspace topology is path-connected. Hence $W$ is path-connected. [/step] [step:Conclude that $U$ has a basis of path-connected open neighborhoods] For the fixed point $x \in U$ and the fixed open neighborhood $V \subset U$, we have constructed an open subset $W \subset U$ such that $x \in W \subset V$ and $W$ is path-connected. Since $x$ and $V$ were arbitrary, every point of $U$ has a neighborhood basis consisting of open path-connected subsets. Therefore $U$ is locally path-connected. [/step]