[proofplan] Fix a base point $a \in U$ and let $C$ be the set of points of $U$ that can be joined to $a$ by a path in $U$. The openness of $U$ lets us join nearby points by straight line segments, which implies that $C$ is open in the relative topology of $U$. The same straight-line argument applied to points in the relative closure of $C$ shows that $C$ is closed in $U$. Since $U$ is connected and $C$ is nonempty, open, and closed in $U$, we must have $C=U$, so every point of $U$ is path-connected to $a$ and hence any two points of $U$ are path-connected to each other. [/proofplan]

[step:Define the path component of a fixed base point] Since $U$ is nonempty, choose a point $a \in U$. Define \begin{align*} C := \{x \in U : \text{there exists a continuous map } \gamma:[0,1]\to U \text{ with } \gamma(0)=a \text{ and } \gamma(1)=x\}. \end{align*} The set $C$ is nonempty because $a \in C$: the constant map $\gamma_a:[0,1]\to U$ defined by $\gamma_a(t)=a$ for all $t \in [0,1]$ is continuous and satisfies \begin{align*} \gamma_a(0)=a \end{align*} and \begin{align*} \gamma_a(1)=a. \end{align*} [/step]

[step:Show that the reachable set is open in $U$]Let $x \in C$. By definition of $C$, there exists a continuous map $\gamma:[0,1]\to U$ such that $\gamma(0)=a$ and $\gamma(1)=x$. For $r>0$, write \begin{align*} B(x,r):=\{z\in\mathbb{R}^n: |z-x|<r\} \end{align*} for the open Euclidean ball of radius $r$ centered at $x$. Since $U$ is open in $\mathbb{R}^n$, there exists $r>0$ such that $B(x,r)\subset U$. We prove that $B(x,r)\subset C$. Let $y \in B(x,r)$. Define the straight-line segment \begin{align*} \sigma_y:[0,1]\to \mathbb{R}^n,\qquad \sigma_y(t)=(1-t)x+ty. \end{align*} For every $t\in[0,1]$, \begin{align*} |\sigma_y(t)-x|=|t(y-x)|=t|y-x|<r, \end{align*} so $\sigma_y([0,1])\subset B(x,r)\subset U$. Define \begin{align*} \eta_y:[0,1]\to U \end{align*} by \begin{align*} \eta_y(t)=\gamma(2t) \quad \text{for } 0\leq t\leq \frac{1}{2}, \end{align*} and \begin{align*} \eta_y(t)=\sigma_y(2t-1) \quad \text{for } \frac{1}{2}\leq t\leq 1. \end{align*} The two formulas agree at $t=\frac{1}{2}$ because \begin{align*} \gamma(1)=x \end{align*} and \begin{align*} \sigma_y(0)=x. \end{align*} Hence the pasting property for continuous maps on the closed subintervals $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ shows that $\eta_y$ is continuous. Also $\eta_y(0)=a$ and $\eta_y(1)=y$, so $y\in C$. Thus $B(x,r)\subset C$, and $C$ is open in the relative topology of $U$.[/step]

[guided]The goal is to prove that once a point $x$ is reachable from $a$, every sufficiently nearby point is also reachable from $a$. Since $x\in C$, there is already a path \begin{align*} \gamma:[0,1]\to U \end{align*} with $\gamma(0)=a$ and $\gamma(1)=x$. For $r>0$, write \begin{align*} B(x,r):=\{z\in\mathbb{R}^n: |z-x|<r\} \end{align*} for the open Euclidean ball of radius $r$ centered at $x$. The only additional ingredient we need is that $U$ is open: because $x\in U$ and $U$ is open in $\mathbb{R}^n$, there exists $r>0$ such that $B(x,r)\subset U$. Now take an arbitrary point $y\in B(x,r)$. We connect $x$ to $y$ by the straight-line segment \begin{align*} \sigma_y:[0,1]\to \mathbb{R}^n,\qquad \sigma_y(t)=(1-t)x+ty. \end{align*} This segment remains inside $B(x,r)$, because for each $t\in[0,1]$, \begin{align*} |\sigma_y(t)-x|=|t(y-x)|=t|y-x|<r. \end{align*} Therefore $\sigma_y([0,1])\subset B(x,r)\subset U$. We concatenate the old path from $a$ to $x$ with this straight-line segment from $x$ to $y$. Define \begin{align*} \eta_y:[0,1]\to U \end{align*} by \begin{align*} \eta_y(t)=\gamma(2t) \quad \text{for } 0\leq t\leq \frac{1}{2}, \end{align*} and \begin{align*} \eta_y(t)=\sigma_y(2t-1) \quad \text{for } \frac{1}{2}\leq t\leq 1. \end{align*} The first formula ends at $x$ because \begin{align*} \gamma(1)=x, \end{align*} and the second formula begins at $x$ because \begin{align*} \sigma_y(0)=x. \end{align*} Thus the two pieces agree at the joining time $t=\frac{1}{2}$. The pasting property for continuous maps on closed subintervals then gives that $\eta_y$ is continuous. Finally, \begin{align*} \eta_y(0)=\gamma(0)=a \end{align*} and \begin{align*} \eta_y(1)=\sigma_y(1)=y. \end{align*} Thus $y\in C$. Since every $y\in B(x,r)$ belongs to $C$, we have found a neighbourhood of $x$ inside $U$ contained in $C$. This proves that $C$ is open in $U$.[/guided]

[step:Show that the reachable set is closed in $U$] Let $x\in U$ belong to the closure $\overline{C}^{\,U}$ of $C$ relative to the [subspace topology](/page/Subspace%20Topology) on $U$. Since $U$ is open, choose $r>0$ such that $B(x,r)\subset U$. Because $x\in \overline{C}^{\,U}$, the relative neighbourhood $B(x,r)\cap U$ intersects $C$. Since $B(x,r)\subset U$, there exists $y\in C\cap B(x,r)$. By definition of $C$, there exists a continuous map $\gamma:[0,1]\to U$ such that $\gamma(0)=a$ and $\gamma(1)=y$. Define \begin{align*} \sigma_x:[0,1]\to \mathbb{R}^n,\qquad \sigma_x(t)=(1-t)y+tx. \end{align*} For every $t\in[0,1]$, \begin{align*} |\sigma_x(t)-x|=|(1-t)(y-x)|=(1-t)|y-x|<r, \end{align*} so $\sigma_x([0,1])\subset B(x,r)\subset U$. Define \begin{align*} \eta_x:[0,1]\to U \end{align*} by \begin{align*} \eta_x(t)=\gamma(2t) \quad \text{for } 0\leq t\leq \frac{1}{2}, \end{align*} and \begin{align*} \eta_x(t)=\sigma_x(2t-1) \quad \text{for } \frac{1}{2}\leq t\leq 1. \end{align*} The two formulas agree at $t=\frac{1}{2}$ because $\gamma(1)=y$ and $\sigma_x(0)=y$. By the pasting property for continuous maps on the closed subintervals $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$, the map $\eta_x$ is continuous. Moreover $\eta_x(0)=a$ and $\eta_x(1)=x$, so $x\in C$. Therefore $C$ contains its relative closure in $U$, so $C$ is closed in $U$. [/step]

[step:Use connectedness to identify the reachable set with all of $U$] We have shown that $C$ is nonempty, open in $U$, and closed in $U$. Since $U$ is connected, the only subsets of $U$ that are both open and closed in the relative topology are $\varnothing$ and $U$. Because $C\neq\varnothing$, it follows that \begin{align*} C=U. \end{align*} Thus for every $x\in U$, there exists a continuous path in $U$ from $a$ to $x$. [/step]

[step:Concatenate paths through the base point to connect arbitrary points] Let $x,y\in U$. Since $C=U$, there exist continuous maps \begin{align*} \gamma_x:[0,1]\to U \end{align*} and \begin{align*} \gamma_y:[0,1]\to U \end{align*} such that $\gamma_x(0)=a$, $\gamma_x(1)=x$, $\gamma_y(0)=a$, and $\gamma_y(1)=y$. Define the reverse path \begin{align*} \widetilde{\gamma}_x:[0,1]\to U,\qquad \widetilde{\gamma}_x(t)=\gamma_x(1-t). \end{align*} Then $\widetilde{\gamma}_x$ is continuous, $\widetilde{\gamma}_x(0)=x$, and $\widetilde{\gamma}_x(1)=a$. Define \begin{align*} \Gamma:[0,1]\to U \end{align*} by \begin{align*} \Gamma(t)=\widetilde{\gamma}_x(2t) \quad \text{for } 0\leq t\leq \frac{1}{2}, \end{align*} and \begin{align*} \Gamma(t)=\gamma_y(2t-1) \quad \text{for } \frac{1}{2}\leq t\leq 1. \end{align*} The two formulas agree at $t=\frac{1}{2}$ because $\widetilde{\gamma}_x(1)=a=\gamma_y(0)$. Therefore $\Gamma$ is continuous by the pasting property. Moreover, \begin{align*} \Gamma(0)=x \end{align*} and \begin{align*} \Gamma(1)=y. \end{align*} Since $x,y\in U$ were arbitrary, $U$ is path connected. [/step]