Every endofunctor on $\mathbf{Set}_{< \aleph_1}$ has an initial algebra. To prove this, here's the key set-theoretical lemma:
Lemma: Let $T$ be an endofunctor on $\mathbf{Set}_{< \aleph_1}$ and let $A_0\subset A_1\subset A_2\subset\dots$ be a sequence of countably infinite sets with union $A$, such that $A_{n+1}\setminus A_n$ is infinite for each $n$. Then the natural map $\operatorname{colim}T(A_n)\to T(A)$ is surjective.
Proof: Suppose $\operatorname{colim}T(A_n)\to T(A)$ is not surjective. That is, there is some element of $T(A)$ that is not in the image of $T(A_n)\to T(A)$ for any $n$. For each $r\in\mathbb{R}$, let $C(r)=\{q\in\mathbb{Q}:q<r\}$, and pick an increasing sequence of rational numbers $(q_n)$ with limit $r$. We can then find a bijection $f:A\to C(r)$ such that $f(A_n)=C(q_n)$ for each $n$ (here we use the assumption that $A_{n+1}\setminus A_n$ is infinite for each $n$). So, there is some element $x_r\in T(C(r))$ which is not in the image of $T(C(q_n))$ for any $n$. It follows that $x_r$ is not in the image of $T(C(s))$ for any $s<r$.
Now let $y_r$ be the image of $x_r$ in $T(\mathbb{Q})$. By construction, $y_r$ is in the image of $T(C(r))\to T(\mathbb{Q})$ but is not in the image of $T(C(s))\to T(\mathbb{Q})$ for any $s<r$. Thus the elements $y_r\in T(\mathbb{Q})$ are all distinct. This is a contradiction, since $T(\mathbb{Q})$ must be countable.
From this lemma, we can deduce the following remarkable fact:
Theorem: Every endofunctor on $\mathbf{Set}_{< \aleph_1}$ preserves sequential colimits.
Proof: Let $T$ be an endofunctor on $\mathbf{Set}_{< \aleph_1}$. We first prove that $A_0\to A_1\to A_2\to\dots$ be a sequential diagram in $\mathbf{Set}_{< \aleph_1}$ with colimit $A$, then the natural map $\operatorname{colim} T(A_n)\to T(A)$ is surjective.
Note first that every surjection in $\mathbf{Set}_{< \aleph_1}$ splits and so $T$ preserves surjections. If the map $A_n\to A$ is surjective for some $n$, the conclusion is thus trivial. So we assume $A_n\to A$ is not surjective for any $n$. In particular, this means $A_n$ is nonempty for sufficiently large $n$, so we may assume $A_n$ is always nonempty. Passing to a subsequence, we may also assume that for each $n$, the image of $A_n$ in $A$ is strictly contained in the image of $A_{n+1}$ in $A$.
Now let $B=A\times\mathbb{N}$ and $B_n=A_n\times\mathbb{N}$. Note that our diagram $(A_n)$ is then a retract of the diagram $(B_n)$, and so the map $\operatorname{colim} T(A_n)\to T(A)$ is a retract of the map $\operatorname{colim} T(B_n)\to T(B)$. Since a retract of a surjection is surjective, it suffices to show $\operatorname{colim} T(B_n)\to T(B)$ is a surjection.
Now let $C_n$ be the image of $B_n$ in $B$. By our previous reductions, we know that each $C_n$ is infinite and that $C_{n+1}\setminus C_n$ is infinite for each $n$. By the Lemma, we conclude that the natural map $\operatorname{colim} T(C_n)\to T(B)$ is surjective. But we have surjections $B_n\to C_n$ compatible with the maps to $B$ for each $n$, and thus surjections $T(B_n)\to T(C_n)$ compatible with the maps to $T(B)$. It follows that $\operatorname{colim} T(B_n)\to T(B)$ is also surjective and thus so is $\operatorname{colim} T(A_n)\to T(A)$.
Note in particular that writing any countable set as a sequential colimit of finite subsets, we see that for any $A$ and any $x\in T(A)$, there is some finite $F\subseteq A$ such that $x$ is in the image of $T(F)\to T(A)$. We now use this to prove that for any sequential diagram $A_0\to A_1\to A_2\to\dots$ with colimit $A$, the natural map $\operatorname{colim} T(A_n)\to T(A)$ is also injective (and thus bijective, so $T$ preserves the colimit).
If $A$ is empty then so is every $A_n$ and the conclusion is trivial, so we assume $A$ is nonempty and then may also assume each $A_n$ is nonempty (if not just drop the finitely many empty terms from the sequence). Now suppose $x,y\in \operatorname{colim} T(A_n)$ have the same image in $T(A)$. There is $n$ such that $x$ and $y$ both lift to $T(A_n)$, and then a finite nonempty subset $F\subseteq A_n$ such that $x$ and $y$ both lift to $T(F)$. Since $F$ is finite, its image in $A_m$ for $m\geq n$ eventually stabilizes. That is, we can choose $m\geq n$ such that the natural map $F'\to A$ is injective, where $F'$ is the image of $F$ in $A_m$. Since $F'$ is nonempty, this injection splits, and so $T(F')\to T(A)$ is injective too. Since $x$ and $y$ have the same image in $T(A)$, their lifts to $T(F')$ must thus be equal. That means their lifts to $T(A_m)$ are equal, so $x$ and $y$ are equal in $\operatorname{colim} T(A_n)$.
It then follows that any endofunctor on $\mathbf{Set}_{< \aleph_1}$ has an initial algebra by Adámek's theorem. Explicitly, we define $A_n=T^n(\emptyset)$ for each $n\in\mathbb{N}$. There are canonical maps $A_n\to A_{n+1}$ given by applying $T^n$ to the unique map $\emptyset\to T(\emptyset)$. Letting $A=\operatorname{colim} A_n$, we also have $A\cong\operatorname{colim} T(A_n)$ since $T(A_n)=A_{n+1}$ and this is compatible with the maps in the system. Since $T$ preserves sequential colimits, the canonical map $A\cong\operatorname{colim} T(A_n)\to T(A)$ is an isomorphism, and its inverse gives a $T$-algebra structure $T(A)\to A$ on $A$, which can be shown to be initial.
