I am self studying Topology from Gemignani's Elementary Topology. Here's the question which I am trying to prove (Exercise 2 on page 127):
Let $X,D$ be a metric space and $\{ s_i \}, i \in I$ be a net in $X$. If every subsequence of $\{ s_i \}$ converges to $x$, then show that $\{ s_i \}$ converges to $x$.
Suppose that $\{ s_i \}$ does not converge to $x$. Now, we're trying to find a subsequence which does not converge to $x$. By the definition, there is a open set $U$ containing $x$ such that for all $i \in I$, $s_j \not\in U$ for some $j \in I$ with $i\le j$. With this, I can easily construct $k : \mathbb{N} \to I$ such that $k$ is monotone and $s_{k_n} \not\in U$ for all $n \in \mathbb{N}$. The only problem I am facing is to find $k$ which satisfies all the properties. I notice that I couldn't even use the "niceness" that metric spaces offer in construction of such a function $k$.
Can someone drop some hints so that I complete this problem? Thanks in advance.
This is an a way a continuous of your previous question and Freakish's answer to it here is still relevant and shows that this statement is wrong: Let any $(x_{\alpha})_{\alpha \in \omega_1}$ be any net into $X$ defined on $\omega_1$, standard order. Then in Gemignani's definition, these net has no subsequences (convergent or otherwise) so vacuously we can say that "all subsequences of $(x_{\alpha})_{\alpha \in \omega_1}$ converge to $x$, whatever $x$ is. If your statement would hold we could conclude that $(x_{\alpha})_{\alpha \in \omega_1}$ converged to $x$, which would almost certainly be false (for all most all $x$ and spaces $(X,d)$).
In short, I'm sceptical. Maybe subsequence has a special meaning on this text (check the index to find the definition?), as @freakish suggested (so not a special subnet).