Is there a Partition of $[0,1]$ into closed, countably infinite sets?

Yes. You can build such a partition by transfinite recursion.

Specifically, we construct a sequence of disjoint closed countably infinite subsets $A_\alpha\subset[0,1]$ where $\alpha$ ranges over the ordinals. Having constructed $A_\beta$ for all $\beta<\alpha$, let $S=[0,1]\setminus\bigcup_{\beta<\alpha}A_\beta$. If $S$ is uncountable, then there is some sequence of distinct elements of $S$ which converges to an element of $S$ (if not, every element of $S$ would be isolated in $S$ and $S$ would be discrete, but there is no uncountable discrete subset of $[0,1]$). Pick such a sequence and let $A_\alpha$ consist of the terms of the sequence together with their limit.

If $S$ is countable, we instead halt the recursion, and put each element of $S$ into a different one of the $A_\beta$ for $\beta<\alpha$ (there must have been uncountably many $\beta<\alpha$ so we can do this, and the sets will remain closed after adding a single point).

Since $[0,1]$ is a set, the recursion must halt eventually, and we obtain a partition of $[0,1]$ into countably infinite closed sets.

It may be worth pointing out that a partition of $[0,1]$ into countably infinite closed sets can be constructed effectively, without using transfinite induction or the axiom of choice.

Let $C$ be the Cantor set. It is well known that we can define a bijection $h:(\frac13,\frac23)\to C.$

If $(a,b)$ is a connected component of $[0,1]\setminus C$ — in other words, $(a,b)$ is one of the "middle thirds" removed from $[0,1]$ in forming the Cantor set — define a bijection $h_{a,b}:(a,b)\to C\cap[2a-b,2b-a].$ No arbitrary choices are needed for this: $[2a-b,2b-a]$ the closed interval of which $(a,b)$ was the deleted middle third, so $C\cap[2a-b,b-2a]$ is similar to the Cantor set, and we can define $h_{a,b}$ as a composition of $h$ with the obvious affine maps: $(a,b)\to(\frac13,\frac23)\to C\to C\cap[2a-b,2b-a].$

Next, I define a map $f:[0,1]\to C.$

If $t\in C,$ let $f(t)=t.$

If $t\in[0,1]\setminus C,$ let $f(t)=h_{a,b}(t),$ where $(a,b)$ is the connected component of $[0,1]\setminus C$ containing $t.$

Finally, for each $x\in C,$ define $A_x=f^{-1}(x).$

Clearly, $\{A_x:x\in C\}$ is a partition of $[0,1],$ and each set $A_x$ is countable, since it contains exactly one point of $C$ and at most one point in each component interval $(a,b)$ of $[0,1]\setminus C.$ A little consideration will show that $A_x$ is also infinite and closed; in fact, it consists of the point $x$ and a sequence of points converging to $x.$