**EDIT**:

As Boaz has pointed out in the comments, there is a mistake in my alleged proof. After thinking about it a bit, I believe it is a genuine mistake (i.e., not quickly fixable). This error does not come in until the end of my argument (Lemma 5), so I will leave the original post up and intact in case the first part of the argument (Lemmas 1-4) can still be useful.

**ORIGINAL POST**:

Every weakly compact cardinal has this property.

I don't know a reference for this, but I can give a proof. The proof is just a generalization of one proof for this fact when $\kappa = \omega$.

**Proposition:** If $\kappa$ is weakly compact, then the generalized Cantor space $2^\kappa$ is homeomorphic to a subspace of every $\kappa$-compact subset $C$ of $2^\kappa$ with $|C| > \kappa$.

**Proof:** I'll try to break it up into bite-sized lemmas:

*Lemma 1*: Suppose $X$ is a Hausdorff space in which an intersection of fewer than $\kappa$ open sets is open. Then every $\kappa$-compact subspace of $X$ is closed. Furthermore, if $X$ is $\kappa$-compact then a subspace $C$ of $X$ is $\kappa$-compact if and only if it is closed.

*Proof*: (In many ways, this is just like the proof for $\kappa = \omega$.) Suppose $A \subseteq X$ is $\kappa$-compact, and fix $x \notin A$. For each $y \in A$, let $U_y$ and $V_y$ be open sets with $x \in U_y$, $y \in V_y$, and $U_y \cap V_y = \emptyset$. The collection $\{V_y : y \in A\}$ covers $A$, so by $\kappa$-compactness some subset $\mathcal{V}$ with $|\mathcal{V}|< \kappa$ still covers $A$. By our assumptions about $X$,
$$\{U_y : V_y \in \mathcal{V}\}$$
is a neighborhood of $x$, and it must be disjoint from $A$. Thus $A$ is closed.

It remains to show that if $X$ is $\kappa$-compact, then every closed $C \subseteq X$ is also. If $\mathcal U$ is an open cover of $C$, then $\mathcal U \cup \{X - C\}$ is an open cover of $X$. Since $X$ is $\kappa$-compact, we may pass to a $<\kappa$-sized subcover. Removing $X-C$ from this subcover if necessary, we obtain a $<\kappa$-sized subset of $\mathcal U$ that covers $C$. $\ $*QED*

*Lemma 2*: Assume $\kappa$ is regular. In the generalized Cantor space $2^\kappa$, any intersection of fewer than $\kappa$ open sets is open.

*Proof:* Using regularity, it is easy to see that this is true for basic open sets. The result follows easily. $\ $*QED*

Putting these two lemmas together with Joseph van Name's answer to your previous question, we obtain:

*Lemma 3*: If $\kappa$ is weakly compact, then $C \subseteq 2^\kappa$ is $\kappa$-compact if and only if it is closed.

*Lemma 4*: If $\kappa$ is regular and $2^{<\kappa} = \kappa$, then every closed $C \subseteq 2^\kappa$ with $|C| > \kappa$ contains a closed subspace with no isolated points.

*Proof*: Fix a closed subset $C$ of $2^\kappa$.

By transfinite recursion, define a sequence of subsets of $2^\kappa$ as follows:
$$C_0 = C$$
$$C_{\alpha+1} = C_\alpha'$$
$$C_{\lambda} = \bigcap_{\alpha < \lambda}C_\alpha \qquad \text{ for limit }\lambda$$
(here $C_\alpha'$ denotes the derived set, or Cantor-Bendixson derivative, of $C_\alpha$). Eventually this sequence stabilizes, and we obtain some $C_\rho \subseteq C$ with $C_\rho = C_\rho'$. Using the fact that $2^\kappa$ has a basis of size $\kappa$ (because $2^{<\kappa} = \kappa$), it is not difficult to show that $\rho < \kappa^+$ and that $|C_{\alpha+1} - C_\alpha| \leq \kappa$ for all $\alpha < \rho$. Because $|C| > \kappa$, it follows that $C_\rho \neq \emptyset$. Furthermore, since $C_\rho = C_\rho'$, $C_\rho$ has no isolated points. $\ $*QED*

*Lemma 5*: Suppose $\kappa$ is regular. Then $2^\kappa$ is the only non-empty, $\kappa$-compact, zero-dimensional Hausdorff space with a basis of size $\kappa$, no isolated points, and in which every intersection of fewer than $\kappa$ open sets is open.

*Proof sketch*: Let $X$ be any space fitting the above description. Let $\{B_\alpha : \alpha < \kappa\}$ be a basis for $X$ consisting of clopen sets.

We now define a map $\varphi$ from the tree $2^{<\kappa}$ into the set of clopen subsets of $X$. Begin by putting $\varphi(\emptyset) = X$. If $\alpha$ is a limit ordinal and $\varphi$ is already defined on $2^{<\alpha}$, then extend the map to $2^\alpha$ by taking intersections:
$$\varphi(f) = \bigcap_{\beta < \alpha}\varphi(f|_\beta) \qquad \text{ for }f \in 2^\alpha.$$
If our map is already defined on $2^\alpha$, then define it on $2^{\alpha+1}$ by splitting each $\varphi(f)$, $f \in 2^\alpha$, into two clopen pieces. Do this in such a way that each piece is nonempty (recall there are no isolated points in $X$). Also, if $\emptyset \neq B_\alpha \cap \varphi(f) \neq \varphi(f)$, then split $\varphi(f)$ into $B_\alpha \cap \varphi(f)$ and its complement.

This tree defines a homeomorphism from $2^\kappa$ to $X$. If $f \in 2^\kappa$, our homeomorphism takes $f$ to the unique element of $\bigcap_{\alpha < \kappa}\varphi(f|_\alpha)$. This intersection is nonempty by $\kappa$-compactness, and it is a singleton by our use of the $B_\alpha$ in the successor stages of our construction. Thus the map is well-defined, and it's not too hard to check that it is a homeomorphism. $\ $*QED*

I admit I left out some details from this last proof. Please feel free to ask for more details if you want them.

Putting everything together, this finishes the proof of the proposition. $\ $**QED**