Let $(F,v)$ be a complete discrete valuation field (normalized) with ring of integers $\mathcal O_F$.
Why cannot exist a valuation ring $A$ of $F$ such that $F\supsetneq A\supsetneq\mathcal O_F$ ? (note that the containments are strict)
Remember that $A$ is a valuation ring of $K$ if for every then $x\in K^\times$ $x\in A\vee x^{-1}\in A$
Let $A$ be any subring of $F$ containing $\mathcal O_F$. Let $\omega$ be a uniformizer for $\mathcal O_F$, so every element $x$ of $F^{\ast}$ can be uniquely represented as $u \omega^n$ for $u \in \mathcal O_F^{\ast}$ and $n \in \mathbb{Z}$. You have $x \in \mathcal O_F$ if and only if $n \geq 0$. If $A$ properly contains $\mathcal O_F$, then there is an $x \in A$ of the form $$x = u \omega^k$$ where $k$ is a negative number. Now $u$ and $\omega$ are in $A$, so this implies $$\omega^{-1} = u^{-1} \omega^{-k}x \in A$$ Since $A$ contains $\mathcal O_F$ and $\omega^{-1}$, you can see that $A$ has to be all of $F$.