[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 $U$.
[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]
[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]
