Let $\mathcal O$ be a normalized discrete valuation ring. This means that there is a surjective valuation: $$v:\text{Frac}\,(\mathcal O)\rightarrow\mathbb Z\cup \left\{\infty\right\}.$$
Now consider a (proper) subring $R\subset\mathcal O$ with the property that $R$ has Krull dimension $>1$. I was wondering what is the group $v(R^\times)\subset\mathbb Z$.
I claim that $v(R^\times)=0$:
If $v(R^\times)\neq 0$ then $v(R^\times)=n\mathbb Z$ for some integer $\mathbb n$. This means that $(R, v_{|_R})$ is a DVR (with a non-surjective valuation) which implies that $R$ is a PID. We conclude that $R$ has Krull dimension $1$ against the assumptions.
Is it correct?
Your argument doesn't work, because you don't necessarily know that $R$ consists of exactly the elements of $\operatorname{Frac}(R)$ whose valuations are nonnegative, and so $R$ need not be a DVR. For a simple example, consider $\mathcal{O}=\mathbb{C}[[x]]$, and let $\alpha\in\mathbb{C}$ be any transcendental element. Then the subring $R=\mathbb{Q}[\alpha,x]$ is $2$-dimensional, but the restriction of the valuation to $\operatorname{Frac}(R)$ is surjective. This doesn't imply $R$ is a DVR because (for instance) $\alpha^{-1}\in \operatorname{Frac}(R)$ has valuation $0$ but is not an element of $R$.