Good morning, I'm trying to solve this exercise:
Could you please verify if my proof looks fine or contains logical gaps/errors? Thank you so much for your help!
My attempt:
Let $\{x_n\}_{n \in \mathbb N}$ be a sequence in $X$ such that $x_n \to \overline x$.
Assume $\{s(x_{n_k})\}_{k \in \mathbb N}$ is a subsequence of $\{s(x_{n})\}_{n \in \mathbb N}$ such that $s(x_{n_k}) \to l$. Because $f$ is continuous $f(x_{n_k}, s(x_{n_k})) \to f(\overline x,l)$. On the other hand, $f(x_n, s(x_n)) = v_0$ for all $n \in \mathbb N$. As such, $f(\overline x,l) = v_0$. Because the equation $f(\overline x,y) = v_0$ has a unique solution $s(\overline x)$. Then $l = s(\overline x)$. To sum up, all cluster points of $\{s(x_{n})\}_{n \in \mathbb N}$ are equal to $s(\overline x)$.
Because $Y$ is compact, $\{s(x_{n})\}_{n \in \mathbb N}$ has at least one cluster point. As such, $s(x_{n}) \to s(\overline x)$ and thus $s$ is continuous.