Assume $C$ and $\mathbb{R}_{\ge 0}^n$ can be (non-strictly) separated by a subspace of dimension $n-1$. Then a normal vector $x$ to that subspace lies in $\mathbb{R}_{\ge 0}^n$; see e.g. here for a proof.

But then by your asummption, there exists $y \in C$ on the same side of this subspace as $\mathbb{R}_{\ge 0}^n$, contradiction.

Assume $C$ and $\mathbb{R}_{\ge 0}^n$ can be (non-strictly) separated by a subspace of dimension $n-1$.
Then a normal vector $x$ to that subspace lies in $\mathbb{R}_{\ge 0}^n$; see e.g. here for a proof.

But then by your assumption, there exists $y \in C$ on the same side of this subspace as $\mathbb{R}_{\ge 0}^n$, contradiction.

Assume $C$ and $\mathbb{R}_{\ge 0}^n$ can be (non-strictly) separated by a subspace of dimension $n-1$. Then a normal vector $x$ to that subspace lies in $\mathbb{R}_{\ge 0}^n$.

But then by your asummption, there exists $y \in C$ on the same side of this subspace as $\mathbb{R}_{\ge 0}^n$, contradiction.

Assume $C$ and $\mathbb{R}_{\ge 0}^n$ can be (non-strictly) separated by a subspace of dimension $n-1$. Then a normal vector $x$ to that subspace lies in $\mathbb{R}_{\ge 0}^n$; see e.g. here for a proof.

But then by your asummption, there exists $y \in C$ on the same side of this subspace as $\mathbb{R}_{\ge 0}^n$, contradiction.