The Hartogs number of $\omega$ is a limit ordinal above $\omega$

Transport of well-orders along an injection: if $\alpha$ is an ordinal and $j:\alpha \to \omega$ is injective, then $S=j[\alpha ]\subseteq \omega$ and $R=\{(j(\xi ),j(\eta )):\xi ,\eta \in \alpha ,\xi \in \eta \}$ well-order $S$, the map $j$ is an order isomorphism of $(\alpha ,\in )$ onto $(S,R)$, and $\mathrm{o}\mathrm{t}(S,R)=\alpha$. Indeed $j$ is a bijection onto $S$; $R$ is irreflexive since $\xi \mathrm{\notin }\xi$; $R$ is transitive and trichotomous because membership on the ordinal $\alpha$ is; and a nonempty $T\subseteq S$ has $j^{-1}[T]$ nonempty with an $\in$-least element $\xi$, whose image $j(\xi )$ is $R$-least in $T$. Since $j(\xi )Rj(\eta )$ holds exactly when $\xi \in \eta$ by injectivity, $j$ preserves and reflects the order and [F10] identifies $\mathrm{o}\mathrm{t}(S,R)$ with $\alpha$.

For every ordinal $\alpha$: $\alpha ⪯\omega$ if and only if $\alpha$ is at most countable. If $j:\alpha \to \omega$ is injective, then $j$ is a bijection of $\alpha$ onto the subset $j[\alpha ]\subseteq \omega$, which is at most countable by [F3], so $\alpha$ is at most countable because the three size classes of [F2] are defined by equinumerosity; conversely, if $\alpha$ is finite then $\alpha \approx m$ for some natural number $m$, and $m\subseteq \omega$ gives an injection $\alpha \to \omega$ by composition, while if $\alpha$ is countably infinite then $\alpha \approx \mathbb{N}=\omega$ gives one directly.

$H=\{\alpha :\alpha \text{ is an ordinal and }\alpha ⪯\omega \}$: if $\alpha =\mathrm{o}\mathrm{t}(S,R)$ for a subset $S\subseteq \omega$ well ordered by $R$, the order isomorphism $\alpha \to S$ is injective into $\omega$; conversely an injection $j:\alpha \to \omega$ produces $S=j[\alpha ]$ and $R$ with $\mathrm{o}\mathrm{t}(S,R)=\alpha$ by step 1.1, so $\alpha$ is one of the order types displayed in [F1].

$H$ is the set of at most countable ordinals and admits no injection into $\omega$: by step 2.1 and step 1.2 an ordinal $\alpha$ lies in $H$ exactly when it is at most countable, and by [F1] the ordinal $H$ itself does not inject into $\omega$.

$\omega \in H$ and $H\mathrm{\ne }0$: the identity is an injection $\omega \to \omega$, so $\omega ⪯\omega$ and hence $\omega \in H$ by step 2.1; therefore $H$ has a member and is not the ordinal $0$, which has none.

$H$ is not a successor ordinal: suppose $H={\alpha }^{+}=\alpha \cup \{\alpha \}$ for an ordinal $\alpha$. Then $\alpha \in H$, so $\alpha ⪯\omega$ by step 2.1, say through an injection $j:\alpha \to \omega$. Define $g$ on the domain ${\alpha }^{+}$ by $g(\beta )=\sigma (j(\beta ))$ for $\beta \in \alpha$ and $g(\alpha )=0$; this is well defined because $\alpha \mathrm{\notin }\alpha$, its values lie in $\omega$ because $\omega$ is inductive, and it is injective: the value $0$ is taken only at $\alpha$ since $\sigma (n)\mathrm{\ne }0$ for every $n$, and on $\alpha$ the map is $\sigma$ composed with the injective $j$, hence injective. So $H={\alpha }^{+}$ would inject into $\omega$, contradicting that $H$ admits no injection into $\omega$ by step 3.1.

$H$ is a limit ordinal with $\omega \in H$, and it is the set of countable ordinals: $H$ is an ordinal by [F1], it is nonzero by step 3.2 and not a successor by step 4.1, so it is a limit ordinal by [F4]; $\omega \in H$ by step 3.2, so $\omega <H$ in the ordinal order of [F5]; the identification $H=\{\alpha :\alpha \text{ is an ordinal and is at most countable}\}$ is step 3.1, and $H$ itself is uncountable because it is the least ordinal that does not inject into $\omega$ and an at most countable ordinal would inject into $\omega$ by step 1.2.

Everything claimed is established: $H=\mathrm{\aleph }(\omega )$ is a limit ordinal with $\omega \in H$, explicitly the set of at most countable ordinals, hence uncountable and not a successor; the argument used only the transport of well-orders along injections, Schröder-free counting, and the Peano properties of $\sigma$, all choice free, so no choice principle is used.