What I really want is this: A sequence $P_0$, $P_1$,... such that each $P_n$ is a countable partition of $\omega_1$, $P_{n+1}$ is a refinement of $P_n$, and such that if $A_n\in P_n$ for all $n$ and $A_{n+1}\subset A_n$ then $\bigcap A_n$ contains at most one point. Just to give it a name, let's say that last condition says that $(P_n)$ is an "injective" sequence of partitions.
This is equivalent to the existence of a injection from $\omega_1$ into $\omega^\omega$, that notation referring to the set of all maps from $\omega$ to itself. Of course such an injection exists, the question is whether there's an "explicit" one.
There's a natural elegant way to construct $P_1$ from $P_0$ and then $P_2$ from $P_1$, etc. Say $P_0=\{I_n:n\in\omega\}$ and assume each $I_n$ is uncountable. Then each $I_n$ is order-isomorphic to $\omega_1$; this isomorphism together with our partition $P_0$ of $\omega_1$ induces a partition $\{I_{n,j}:j\in\omega\}$ of $I_n$, and now we let $P_1=\{I_{n_0,n_1}:n_0,n_1\in\omega\}$. Now we use $P_0$ again to define a partition of each $I_{n_1,n_1}$, etc.
A $P_0$ such that this construction gives an "injective" sequence $P_0,P_1,\dots$ would be cool. Alas what strikes me as the natural choice of $P_0$ fails. (For a minute I thought that the fact that the construction fails for one $P_0$ implies that it fails in general. When I think about the details this becomes less clear; if you feel like asserting that it must fail in general because it fails for one $P_0$ please include a little explanation.)
An explicit injective sequence of partitions by some other construction would be fine...
What you are attempting is not possible in general. Here is a counterexample: It is consistent (modulo large cardinals) that dependent choice holds, but full choice fails, and the club filter is an ultrafilter on $\omega_1$. Since the club filter is countably complete, it follows that in such a model, if we partition $\omega_1$ as $\bigcup_n B_n$, then exactly one $B_n$ contains a club. If we have a sequence $P_0,P_1,\dots$ of partitions of $\omega_1$ as you want, for each $n$ there is exactly one element of $P_n$ that contains a club. Call it $A_n$, and note that $A_{n+1}\subseteq A_n$ for all $n$. But then the intersection of the $A_n$ also contains a club. (A quick proof of the last claim goes by noting that the assumptions ensure that $\omega_1$ is regular and that we can simultaneously pick clubs $C_n\subseteq A_n$ from each $n$. But then one easily checks that $\bigcap_n C_n$ is also club.)