Finer than the Fréchet filter implies that $\mathcal{U}$ is free, otherwise there's be some unique $m \in \cap \mathcal{U}$. Then the ultrafilter space would not be Hausdorff, so "ugly" (we cannot separate $m$ from $\omega$). For a free ultrafilter the resulting space will be Hausdorff (separate $m$ from $\omega$ by $\{m\}$ and $\{\omega\} \cup \mathbb{N}\setminus \{m\}$, the last set being in the Fréchet filter) , even normal (Any Hausdorff set with only one non-isolated point). Hence the condition ,I think, it's to have a "nice" ultrafilter space.
The convergence remark is rather easy to see: suppose $(x_n)$ is a sequence in $X$ where the range $R = \{x_n: n \in \mathbb{N}\}$ is infinite.
The only way a the sequence could converge to $p \in \mathbb{N}$ is when all $x_n$ from some $N$ onwards would equal $p$: definition of convergence applied to the open set $\{p\}$. But this would make $R = \{x_1, \ldots, x_N\}$ a finite set. On the other hand, divide $R$ into two disjoint infinite sets $R_1$ and $R_2$. Then either $R_1$ or its complement $\mathbb{N}\setminus R_1$ lies in $\mathcal{U}$, by a standard property of ultrafilters. If $R_1 \in \mathcal{U}$, the open set neighbourhood $R_1 \cup \{\omega\}$ misses infinitely many terms of the sequence, so the sequence cannot be eventually in his open set. The same argument applies to the open set $\mathbb{N}\setminus R_1 \cup\{\omega\}$ in the other case. This makes $X$ a basic example: $\omega \in \overline{\mathbb{N}}$ but no sequence from $\mathbb{N}$ can converge to $\omega$, which shows sequences are not enough to describe the topology even in nice (normal and Hausdorff) spaces like this $X$.
My question is as follows: why there is the need to assume that $\mathfrak U$ is an ultrafilter that is finer than the Fréchet filter. This seems unnecessary to me since the statement is true for any ultrafilter.