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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?

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    What role does the set $U$ play in your proof? And I don't really see how you're getting a contradiction. – 2017-01-26
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    $U$ is an open set as defined to satisfy the conditions in the first couple of lines. The contradiction is there because if $\gamma(t')\in B(\gamma(t'))$ and $B(\gamma(t'))\subset E$ then $\gamma(t')$ must be in $E$. – 2017-01-26
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    How do you know that there is a ball $B(\gamma(t^{\prime}))\subset E$ though? And you never use the set $U$ in the proof. – 2017-01-26
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    That's the thing. I want to explicitly argue in (i) that since $E\subset U$, then $B(\gamma(t'))\subset U$. There must be some simple way to do that. Recursively speaking from the definition given above for connected sets, $U$ cannot be comprised of disjoint sets. – 2017-01-26

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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.

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    I think that I can immediately use the definition of connectedness given in my question to conclude that $A\cap B=\emptyset$ (or $U\cap V=\emptyset$) is impossible since $E$ must be a subset of $U$ or $V$. – 2017-01-26
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    $A\cap B=\emptyset$ is possible if $B=\emptyset$, and indeed this is the point of the proof. – 2017-01-26
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    But yes, it needs to be shown that $A$ and $B$ are open to be able to apply the definition. – 2017-01-26
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    I think I can just say that since $E$ is open, $A\cup B=E$, and that since if a finite union of sets is open implies that the union consists of open sets, both $A$ and $B$ must be open. – 2017-01-26
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    That's not true, unfortunately. For instance, $(0,1)=(0,\frac{1}{2}]\cup(\frac{1}{2},1)$. – 2017-01-26
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    I'm not sure how to prove that $A$ and $B$ are open using the path-connectedness of open balls. But can I validly say in my original proof that $U$ must not be comprised of any disjoint sets (by induction, say)? – 2017-01-26
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    Suppose that $E$ is connected. Then if disjoint sets $ U,V$ are such that $E\subset U\cup V$, then, WLOG, $V$ is empty. Then $E\subset U$. Suppose that $U$ is comprised of disjoint sets $U_1, V_1$. Then, WLOG, $E\subset U_1$. Suppose that $U_n$ is comprised of disjoint sets $U_{n+1}, V_{n+1}$. As $n\to \infty$, by the Nested Subsets Lemma, $E \subset U_\infty$, where $U_\infty$ is a singleton set. Since singletons are closed, $E$ is closed or empty. But this is a contradiction, thus $U$ must not be comprised of disjoint sets. @carmichael561 – 2017-01-26
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    Here's how to prove that $A$ is open: suppose that $y\in A$. Since $E$ is open, there is an open ball $D$ such that $y\in D\subset E$. If $z\in D$ then there is a path from $y$ to $z$ since $D$ is path connected. Since there is also a path from $x$ to $y$, by gluing these paths together we get a path from $x$ to $z$. Therefore $D\subset A$, so $A$ is open. – 2017-01-26
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    What if $z\not\in D$? – 2017-01-26
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    To show that $A$ is open, you have to show that every $y\in A$ is contained in an open ball which is contained in $A$. So there's no reason to consider $z\not\in D$. – 2017-01-26