First, your question is somewhat misleading:
The condition that you're asking for is equivalent to connectedness. In particular, a space $X$ is disconnected if there exist nonempty open sets $U$ and $V$ so that $X=U\cup V$ and $U\cap V=\emptyset$. Moreover, if $W$ is a nontrivial clopen set, then $W$ and $W^c$ are both open, $X=W\cup W^c$, and $W\cap W^c=\emptyset$. Therefore, they form a separation of $X$. Hence, by asking for a nontrivial clopen set, you are asking for a separation and using the definition of disconnected.
What I think that you are asking is to avoid the implication path connected implies connected. So, to prove that $\mathbb{R}^n$ is connected without going through path connected.
Suppose that $\mathbb{R}^n\setminus\{0\}$ has a nontrivial clopen set $X$. Then, since $\mathbb{R}^n\setminus\{0\}$ has the induced topology, there is an open set $U$ in $\mathbb{R}^n$ and a closed set $D$ in $\mathbb{R}^n$ so that $U\cap \mathbb{R}^n\setminus\{0\}=X$ and $D\cap\mathbb{R}^n\setminus\{0\}=X$. We note that $U$ and $D$ only differ by a single point $\{0\}$ since their intersection with $\mathbb{R}^n\setminus\{0\}$ is the same. Moreover, the two sets must differ because $\mathbb{R}^n$ is connected and has non nontrivial clopen sets.
Let $x\in X$ and let $y\not\in X$. By perturbing $x$ and $y$ slightly, we may assume that the line $\ell$ though $x$ and $y$ does not include $0$ (since $X$ is clopen, there are small neighborhoods around $x$ in $X$ and around $y$ in in $X^c$). Then the induced topology from $\mathbb{R}^n$ to $\ell$ produces the standard topology on $\ell\simeq\mathbb{R}$. We observe that $U\cap \ell=D\cap \ell$, since their only difference is at $0$. Since $x\in U\cap\ell$ and $y\not\in U\cap \ell$, this defines a nontrivial clopen set in $\mathbb{R}$, which contradicts the fact that $\mathbb{R}$ is connected.
