The link to the paper $I^K$Convergence by Sleziak and Macaj
Example $3.15:$ Let us recall that $ω_1$ denotes the first uncountable ordinal with the usual ordering. Let $X$ be the topological space on the set $ω_1 ∪ \{ω_1\}$ with the topology such that all points different from $ω_1$ are isolated and the base at the point $ω_1$ consists of all sets $U_α = \{β ∈ X; β > α\}$ for $α < ω_1$. Notice that if $C ⊆ ω_1$ is a set such that $ω_1 ∈ \bar C$, then $|C| = ℵ_1$.
Now let $I$ be an admissible ideal on $\mathbb N$ and let a function $f : \mathbb N → X$ be $I$-convergent to $ω_1$. We will show that then there exists $M ∈ F(I)$ such that $f(x) = ω_1$ for each $x ∈ M$, that is, $f|_M$ is constant. Clearly, this implies that $f$ is $I∗$ -convergent.For the sake of contradiction, suppose that each set $M ∈ F(I)$ contains some point $m$ such that $f(m)\neq ω_1$. Since $f^{−1} (U) ∈ F(I)$, for any neighborhood $U$ of $ω_1$ in $X$ there exists $m ∈ \mathbb N$ with $f(m) ∈ U \ \{ω_1\}$. Therefore for the set $C = \{m ∈ \mathbb N; f(m)\neq ω_1\}$ we have $ω_1 ∈ f[C]$. Since $f[C] ⊆ ω_1$ and it is a countable set contained in $ω_1$, this is a contradiction. Now, by choosing an ideal $I$ which does not have the additive property $AP(I, Fin)$ we obtain the desired counterexample.
$1)$"Notice that if $C ⊆ ω_1$ is a set such that $ω_1 ∈ \bar C$, then $|C| = ℵ_1$": Is this another theorem or am I supposed to derived from here?In the later case I'll really need a little hint.
$2)$ When we say $C\subset \omega_1$ we mean that $C$ is family(may be singleton even) of countable ordinals, is it$?$The space $X$ here something like $\mathbb R \bigcup \{\infty\}$ if we denoted $\mathbb R$ by $\infty$. Every member of $\mathbb R$ itself is a finite number but their upper bound is $\infty$ which is nothing finite.Similarly every member of $\omega_1$ is a countable ordinal but $\omega_1$ itself is not although it is the upper limit.
Thank you.
Suppose that $C\subseteq\omega_1$ is countable; then $\sup C<\omega_1$. Let $\alpha=\sup C$; then $U_\alpha$ is an open nbhd of the point $\omega_1$ that is disjoint from $C$, so $\omega_1\notin\operatorname{cl}C$.
Your understanding in your second question is correct, though $\Bbb N\cup\{\infty\}$ is a better analogue. Just as $\Bbb N\cup\{\infty\}$ is the one-point compactification of $\Bbb N$, this space $X$ is the one-point Lindelöfization of a discrete space of cardinality $\omega_1$.