I'm proving this theorem:
Let $X,Y$ be metric spaces. Then a function $f:X \to Y$ is continuous at $x$ if and only if $f$ is sequentially continuous at $x$.
My questions:
Does my proof look fine or contains logical gaps/errors? Any suggestion is greatly appreciated.
Is it possible to utilize a weaker form of Axiom of Choice?
Thank you for your help!
My attempt:
"$\Longrightarrow$": Assume that $f:X \to Y$ is continuous at $x$ and that $(x_k)_{k \in \mathbb N}$ converges to $x$. For every neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ such that $f(U) \subseteq V$. On the other hand, all most all $x_k$'s are in $U$, so all most all $f(x_k)$'s are in $V$. Hence $(f(x_k))_{k \in \mathbb N}$ converges to $f(x)$ and thus $f$ is sequentially continuous at $x$.
"$\Longleftarrow$": Assume the contrary that $f$ is sequentially continuous at $\overline x$ but discontinuous at $\overline x$. Then there is a neighborhood $V$ of $f(\overline x)$ such that $f[\mathbb B(\overline x,1/k)] \not \subseteq V$ for all $k$. We define a sequence $(A_k)_{k \in \mathbb N}$ by $A_k = \{x \in \mathbb B(\overline x,1/k) \mid f(x) \notin V\}$. By assumption, $A_k \neq \emptyset$ for all $k$. By Axiom of Choice, there exists a sequence $(x_k)_{k \in \mathbb N}$ such that $x_k \in A_k$ for all $k$. By construction, $(x_k)_{k \in \mathbb N}$ converges to $\overline x$. On the other hand, $(f(x_k))_{k \in \mathbb N}$ does not converge to $f(\overline x)$ because all of $f(x_k)$'s do not belong to the neighborhood $V$ of $f(\overline x)$. This is a contradiction. As a result, if $f$ is sequentially continuous at $x$, then $f$ continuous at $x$.
