Let $\displaystyle x \in O$ where $\displaystyle O$ is open. Prove that if $\displaystyle x_{n} \rightarrow x$ then all but a finite number of terms of $\displaystyle (x_{n})$ must be contained in $\displaystyle O$.
As Moo said that is literally the definition of $\displaystyle x_n\to x$. If this is a metric space and that is not how it's defined, merely note that since $\displaystyle x\in O$ there exists some $\displaystyle \varepsilon>0$ such that $\displaystyle B_{\varepsilon}(x)\subseteq O$. But, by definition there exists some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\leqslant n\implies d(x_n,x)<\varepsilon\implies x\in B_{\varepsilon}(x)\implies x\in O$