I tried to prove this in the Kunen's set theory:
Let $P$ be a countable non-atomic partial order. Show that there is a dense embedding from $T = \{p\in\operatorname{Fn}(\omega,\omega) :\operatorname{dom} p\in\omega\}$
At first I tried as follows: Let $A_0$ be a maximal antichain in $P$. Since $P$ is non-atomic, $A_0$ is countable. I shall construct the function $i$ from $T$ to $P$. Let $\{q_k : k<\omega\}$ be an enumeration of $A_0$ then define $i(\{(0,k)\}) = q_k$.
Let assume that $i(p)$ with $\operatorname{dom} p = n$ is given. If $A=\{q_k : k<\omega\}$ is a countable maximal antichain lies below $i(p)$ then define $i(p\cup\{(n, k)\}) = q_k$. It gives the order-preserving function $i$ from $T$ to $P$.
However I realized that my attempt does not work in some cases. For example, above construction need not always work in the case $$P = T \cup \{p\in\operatorname{Fn}(\omega+\omega,\omega) : \omega\subset\operatorname{dom} p \in \omega+\omega\>\text{ and }\>\forall n<\omega :p(n)=0\}$$ because we can take each antichain appeared in the construction as subset of $T$. I don't know how to avoid such problem nor find an other proof of it. I would be appreciated any help.
Here is a small modification of your construction which explicitly ensures that the embedding we build is dense.
Let $P=\{p_n;n\in\omega\}$ be an enumeration of $P$. We build the embedding $i\colon T\to P$ level by level. At the start, we map the top condition of $T$ to the top condition of $P$.
To define $i$ on the nodes of length $1$ we let $A_\emptyset$ be an infinite maximal antichain in $P$ (such a thing exists because $P$ is atomless) such that some $p\in A_\emptyset$ is below $p_0$. We then enumerate $A_0$ and map the $k$-th node of length $1$ in $T$ to the $k$-th element of $A_\emptyset$.
Proceeding by induction, assume that we have defined $i(s)$ for every $s\in T$ of length at most $n$ and consider an $s\in T$ of length $n$. To define $i$ on the immediate successors of $s$ let $A_s$ be an infinite maximal antichain in $P$ below $i(s)$ such that, if $p_n$ is compatible with $i(s)$, some element of $A_s$ is below $p_n$. We then enumerate $A_s$ and map the successors of $s$ to the conditions in $A_s$.
It is now clear that the embedding $i$ is dense; the image of every level of $T$ is a maximal antichain in $P$, so any $p_n$ will be compatible with some condition in the image of the $n$-th level, which means that the image of some node on the $(n+1)$-st level will get below $p_n$.