Here's how I proceed:
(i) Show that, since E is connected, for any open disjoint sets $U, V\subseteq \mathbb{R}^n$, such that $E\subseteq U\cup V$, either $U\cap E$ or $V\cap U$ is empty. WLOG, $V$ is empty, so $E\subseteq U$.
(ii) Suppose that $E$ is not connected. Then, for a continuous function $\gamma: [0,1]\to E$ and any start and end points $x_1, x_2$, there exists a $t'\in [0,1]$ such that $\gamma(t')\not\in E$.
(iii) Since $E$ is open, for every $x\in E$, there exists an open ball $B$ such that $B\subset E$. Hence, $\gamma(t')\in B(\gamma(t'))\subset E$, which implies that $\gamma(t')\in E$, a contradiction. Hence, $E$ is path-connected.
However, I'm a bit concerned about part (i). There I show that $E$ is a subset of an open set. It is understood that $U$ cannot be comprised of any disjoint sets. But I feel that something needs to be added to part (i) to emphasize this. Is there a concept which defines a set as being a "whole" set in analysis?
I will assume that $E$ is non-empty to avoid worrying about whether the empty set should be regarded as path connected or not.
Fix $x\in E$, and let $A$ be the set of all $y\in E$ for which there is a path from $x$ to $y$ in $E$. Note that $A$ is non-empty since $x\in A$. Also let $B$ be the set of all $y\in E$ for which there is no path from $x$ to $y$ in $E$.
We have $E=A\cup B$ and $A\cap B=\emptyset$, so to show that $A=E$ it is enough to show that both $A$ and $B$ are open, for then $B$ must be empty because $E$ is connected.
And to show that $A$ and $B$ are open, use the fact that $E$ is open together with the fact that balls in $\mathbb{R}^n$ are path connected.