I am having some trouble with this demonstration.
I know that the definition of continuous preference relation is: $$\succsim \text{is continuous iff } \forall \text{ pair of sequences } \{x^n\}_{n>0} \text{ and } \{y^n\}_{n>0} \text{ such that } x^n\succsim y^n \ \forall n, \ x^n \rightarrow x, \ y^n \rightarrow y, \text{ we have } x\succsim y$$
I think I need to use this definition of closed set: $$S \text{ is closed iff every convergent sequence } \{a_n\}_n \text{ converges in S}$$
Using the definition of upper contour set $U_{x^0}=\{x \in X: x\succsim x^0\}$, maybe I need to use the fact that $\{x^n\}_{n>0} \in U_{x^0}$ and converges to $x \in U_{x^0}$. But I don't know if this is the correct approach.
Thank you very much.
As you correctly said, pick a sequence $(x^n)_{n>0} \in U_{x^0}^{\mathbb{N}}$ that converges to some $x$. What we have to show is that $x \in U_{x^0}$, which means $ x\succsim x^0$.
Now, since you have continuity (note: it is usually known as sequential continuity of a preference relation, to distinguish from the Debreu continuity, the algebraic continuity and the Archimedean continuity), just consider as $(y^n)_{n>0}$ the constant sequence of $x^0$. Then by sequential continuity you get that the inequality is preserved in the limit, that is $ x\succsim x^0$, and so $x \in U_{x^0}$.