$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$Let $\R_-:=(-\infty,0]$ and $\R_+:=[0,\infty)$. The desired statement is equivalent to the following:
Suppose that \begin{equation*} \forall x\in X\ \exists i\in[n]:=\{1,\dots,n\}\ x_i>0. \tag{1}\label{1} \end{equation*} Then there is some $\nu\in\R_+^n$ such that \begin{equation*} \nu\cdot X:=\{\nu\cdot x\colon x\in X\}\subseteq\R_+\quad\text{and}\quad \nu\cdot X\ne\{0\}. \tag{2}\label{2} \end{equation*}
Proof: Condition \eqref{1} means that $X\cap\R_-^n=\emptyset$. So, by Theorem 11.3 in Rockafellar (1970), there is some $\la^1\in\R^n\setminus\{0\}$ such that $\la^1\cdot X\subseteq\R_+$ and $\la^1\cdot\R_-^n\subseteq\R_-$. Also, $\la^1\cdot\R_-^n\subseteq\R_-$ means that $\la^1\in\R_+^n$. Thus, \begin{equation*} \text{$\la^1\in\R_+^n\setminus\{0\}$ and $\la^1\cdot X\subseteq\R_+$.} \tag{3}\label{3} \end{equation*}
So, either \eqref{2} holds with $\la^1$ in place of $\nu$ or
$\la^1\cdot X=\{0\}$.
So, without loss of generality (wlog), $\la^1\cdot X=\{0\}$. That is, we have
\begin{equation*}
X\subseteq V^1:=\{x\in\R^n\colon\la^1\cdot x=0\}.
\end{equation*}
The condition $\la^1\in\R_+^n\setminus\{0\}$ in \eqref{3} implies that wlog for some $k^1\in[n]$ and all $i\in[n]$
\begin{equation*}
\la^1_i>0\text{ if }i\le k^1\quad \text{and}\quad \la^1_i=0\text{ if }i>k^1. \tag{4}\label{4}
\end{equation*}
Now condition \eqref{1} implies $X\cap V^1_-=\emptyset$, where $V^1_-:=V^1\cap\R_-^n$. Applying now Theorem 11.3 in Rockafellar (1970) to the Euclidean space $V^1$, we see that there is some $\mu^1\in V^1\setminus\{0\}$ such that \begin{equation*} \mu^1\cdot X\subseteq\R_+ \end{equation*} and $\mu^1\cdot(V^1\cap\R_-^n)\subseteq\R_-$. The latter condition means that for all $x\in\R^n$ we have \begin{equation*} (\la^1\cdot x=0\ \&\ x\in\R_-^n)\implies\mu^1\cdot x\le0 \end{equation*} or, equivalently, \begin{equation*} (\la^1\cdot x=0\ \&\ x\in\R_+^n)\implies\mu^1\cdot x\ge0. \tag{5}\label{5} \end{equation*} Substituting the $i$th standard basis vector $e_i$ for $x$ in \eqref{5}, and recalling \eqref{4}, we get $\mu^1_i=\mu^1\cdot e_i\ge0$ for $i>k$.
So, wlog there is some real $t^1>0$ such that for
\begin{equation*}
\la^2:=\la^1+t^1\mu^1 \tag{6}\label{6}
\end{equation*}
and some integer $k^2\in[k^1,n]$ we have
\begin{equation*}
\la^2_i>0\text{ if }i\le k^2\quad \text{and}\quad \la^2_i=0\text{ if }i>k^2. \tag{4a}\label{4a}
\end{equation*}
Also, by \eqref{6}, $\la^2\cdot X\subseteq\la^1\cdot X+t^1\mu^1\cdot X\subseteq\R_+$. So, either \eqref{2} holds with $\la^2$ in place of $\nu$ or
$\la^2\cdot X=\{0\}$ and hence $\mu^1\cdot X=\{0\}$.
So, wlog, $\la^2\cdot X=\{0\}=\mu^1\cdot X$. So,
\begin{equation*}
X\subseteq V^2:=\{x\in V^1\colon\la^2\cdot x=0\}
=\{x\in V^1\colon\mu^1\cdot x=0\}\subsetneq V^1;
\end{equation*}
the latter strict inclusion follows because $\mu^1\in V^1\setminus\{0\}$.
Continuing thus, in $n$ similar steps wlog we will get $X\subseteq V^n:=\{0\}$, which contradicts \eqref{1}. $\quad\Box$.