Totally disconnected subspaces This question is motivated by this one, where no simple solution within ZFC seems to exist. Let me ask a weaker question then.
Suppose that $K$ is a compact, Hausdorff, non-metrizable space. Does it contain a closed, non-metrizable, totally disconnected subspace?
Note that the cardinality of the (consistent) counterexample in Mathieu Baillif's answer is ludicrously large ($=\omega_{\omega_1}$). Hence my question.
 A: EDIT (24th feb. 2015): I added a paragraph at the end to answer a question of Tomek Kania.
As I wrote in a comment above, the Nyikos space $X$ I used in my answer to the question cited above gives a consistent counter-example (when taking the one-point compactification).
This space is built with the help of the axiom $\diamondsuit$. I just noticed that a construction can be made using only the axiom $\clubsuit_C$ ("club for clubs") which is much weaker. In particular, $\clubsuit_C$ is compatible with MA + $\neg$CH. I am 100% sure that this construction has been made before, by Nyikos probably, but I did not find it in print, so here is a sketch of how it works.
Let $\Lambda$ be set of limit ordinals in $\omega_1$. The axiom $\clubsuit_C$ says:
$\clubsuit_C$: There exists $\langle S_\lambda:\lambda\in\Lambda\rangle$ such that $S_\lambda$ is an increasing $\omega$-sequence converging to $\lambda$, with the property that if $C\subset\omega_1$ is closed and unbounded (abbreviated by "club"), there is some $\lambda\in\Lambda$ with $S_\lambda\subset C$.
The idea is the following. 
The underlying set of the space $Z$ is $\omega_1\times[0,1]$ (in the end we take the one-point compactification and basically add $\{\omega_1\}\times\{0\}$).
Write $S_\lambda=\langle s^\lambda_1, s^\lambda_2,\dots\rangle$ in increasing order. We define the topology by induction such that $Z_\lambda=\lambda\times[0,1]$ is homeomorphic to a closed subspace of $\lambda\times[0,1]$ with the product topology. Thus, $Z_{\lambda+1}$ is compact metrizable. Moreover, the homeomorphism preserves the fibers.
The way the topology on $Z_{\lambda+1}$ (for limit $\lambda$) is defined is summarized in this picture.

We arrange such that for any $\omega$-sequence $x_n$ in $[0,1]$, the
sequence $\langle s_n^\lambda,x_n\rangle$ has all of $\{\lambda\}\times[0,1]$ in its closure. 
If $\lambda$ is limit, the topology of $Z_\lambda$ is just the one given by the increasing union, and $Z_{\lambda+2}$ is the disjoint union of $Z_{\lambda+1}$ and $\{\lambda+1\}\times[0,1]$.
Taking $Z=\cup_{\lambda<\omega_1}Z_\lambda$, we obtain a countably compact first countable locally compact and locally metrizable space. If $E\subset Z$ is closed non-metrizable, then it is not contained in any $Z_\lambda$ and one sees easily that it must intersect the fibers in a closed and unbounded subset $C$ of $\omega_1$. But then, by $\clubsuit_C$, $E$ must contain a sequence $\langle s_n^\lambda,x_n\rangle$ and thus all of $\{\lambda\}\times[0,1]$ for some $\lambda$. Thus $E$ is not totally disconnected.
The one point compactification of $Z$ has the same property.
I don't know whether this type of things can be done in ZFC alone. The PFA (and weaker axioms as well) impedes such a construction with a first countable countably compact $Z$ because it imposes the presence of a copy of $\omega_1$ in it. And there is a complete forest of results about compact spaces under PFA which might imply a positive answer, but I am too ignorant to know.
Notice that using $Z$ instead of $X$ and any cardinal $\kappa> 2^\omega$ of cofinality $\omega_1$ instead of $\omega_{\omega_1}$ in my answer here we obtain a counter-example for that question as well under $\clubsuit_C$.

Tomek Kania asked whether there is a hereditarily separable such example. After some browsing of the litterature, I found one which exists under $\diamondsuit$. It is described on pages 652--655 of Nyikos's paper "The theory of nonmetrizable manifolds" in the Handbook of set-theoretic topology. 
The idea (due to Rudin-Zenor) is to use $CH+\clubsuit$ (which is equivalent to $\diamondsuit$) to obtain a $2$-dimensional nonmetrizable hereditarily separable manifold $M=\cup_{\alpha<\omega_1}M_\alpha$ such that $M_\alpha$ is a metrizable dense submanifold of $M$. ($M$ is in fact perfectly normal and countably compact.) Moreover, $M_{\alpha+1}$ is obtained by "insterting" a half open interval $[0,1)$ inside $M_\alpha$ in so that the following holds (see p.654):
Claim: If $F$ is a closed non-Lindelöf subset of $M$, then $F$ contains a copy of $[0,1)$, and is therefore not totally disconnected.
To finish, notice that if $F$ is non-metrizable, then it is non-Lindelöf. 
Then, take the one point compactification of $M$ to obtain the desired example. 
I know that the original Rudin-Zenor construction of a perfectly normal nonmetrizable (hereditarily separable) manifold was done with $CH$ alone, I'd have to check whether the above claim also holds in their construction.
A: The following is Question 374 in Open Problems in Topology II, and it is attributed to S. Todorcevic:

Is it consistent that every nonmetrizable perfect compactum contains a closed
  subspace with uncountably many clopen sets?

By the context I think "perfect compactum" means "perfecly normal compact space". Note that a closed, non-metrizable, totally disconnected subspace of a compact space must have uncountably many clopen sets. So it seems that either the consistency of a positive answer to the OP´s question is an open problem (at least in 2007), or there is a real counterexample that we are missing (note that the OP´s question does not require the space to be $T_6$ and requires the subspace to be zero-dimensional, so it should be easier to find a counterexample).
