Let $\gamma:[a;b]\to\mathbb{R}^2$ is jordan curve. Proof $\mathbb{R}^2\setminus\mathrm{Im}(\gamma)$ is not connected.
Here, Jordan curve is a closed simple continuous plane curve.
I think because $\mathbb{R}^2\setminus\mathrm{Im}(\gamma)$ is union of two open sets but i don't show that?
You are right: since $\mathbb{R}^2\setminus\operatorname{Im}(\gamma)$ is the union of two (disjoint) open sets, it cannot be connected. To be more precise: if $X$ is a topological space and if $A_1$ and $A_2$ are disjoint non-empty subsets of $X$, then $A_1\cup A_2$ is not connected (with respect to the induced topology), since $A_1\cup A_2$ is the union of two open disjoint subsets, none of which is empty.