List the subsets of (for example) $\{0,1,2,3,4,5,6,7,8,9\}.$ You have only finitesly many. Then make a second copy of that list and for each set $S$ in the list add $S\cup\{10\}$. That's another finitely many that you can append to the list you already had. Then make a second copy of this new longer finite list and for each set $S$ in that list, add the set $S\cup\{11\}.$ That's another finitely many that you can append to the list you already had. And so on. You get a list of all finite subsets of $\omega$, and each member of your list has only finitely many others before it.
Let's try this starting with the empty set $\varnothing$ rather than with $\{0,1,2,3,4,5,6,7,8,9\}:$
$$
\begin{array}{lrl}
& 1 & \varnothing \\
\hline
\text{Copy the list above and add the next member:} & 2 & \{0\} \\
\hline
\text{Copy the list of 2 sets above and add the next member:} & 3 & \{1\} \\
& 4 & \{0,1\} \\
\hline
\text{Copy the list of 4 sets above and add the next member:} & 5 & \{2\} \\
& 6 & \{0,2\} \\
& 7 & \{1,2\} \\
& 8 & \{0,1,2\} \\
\hline
\text{Copy the list of 8 sets above and add the next member:} & 9 & \{3\} \\
& 10 & \{0,3\} \\
& 11 & \{1,3\} \\
& 12 & \{0,1,3\} \\
& 13 & \{2,3\} \\
& 14 & \{0,2,3\} \\
& 15 & \{1,2,3\} \\
& 16 & \{0,1,2,3\} \\
\hline
\text{Copy the list of 16 sets above and add the next member:} & & \cdots \cdots \cdots
\end{array}
$$