Is there a non-locally compact Hausdorff space in which all infinite compact sets (of which there is at least one) have uncountable interiors? Here is the background material from which I am working:


*

*The Cantor set is an uncountable compact Hausdorff subspace of $\mathbb{R}$ with empty interior.

*In a locally compact Hausdorff space with no isolated points, each countable set has empty interior.

*The rational numbers with the subspace topology is a non-locally compact Hausdorff space in which all compact sets have empty interior.


I am trying to find a non-locally compact Hausdorff space in which there are infinite compact sets, and where all of the infinite compact sets have uncountable interiors. I am guessing the example will be an exotic function space.
I first posed this question without specifying that there should be at least one infinite compact set, and this was solved by Stefan H. on this site. I then posed this updated question, which was solved by Dejan Govc. 
 A: It seems the following. 
At last you obtain a positive answer, because of the following
Lemma. Each infinite locally compact Hausdorff space $X$ contains an infinite closed subspace with a countable interior. 
So if you have an infinite compact subspace $X$ of a Hausdorff space $Z$ then  $X$ contains an infinite closed subspace with a countable interior not only in $Z$, but even in $X$.
Proof of Lemma. Suppose the converse. Let $Y=\{y_n:n\in\omega\}$ be an arbitrary countable infinite subset of $X$. Then the set $\overline Y$ is uncountable. Let $x\in \overline Y\backslash Y$ be an arbitrary point. Since the space $X$ is Hausdorff, for each number $n\in\omega$ there exist disjoint open neighborhoods $U_n$ and $V_n$ of the points $y_n$ and $x$ respectively. The set $\overline Y\backslash\bigcup U_n $ is a closed set with an empty interior and therefore it is finite. Then the set $\overline Y\cap \bigcap V_n\subset \overline Y\backslash\bigcup U_n$ is finite too. Thus the point $x$ has a countable pseudocharacter $\psi(x,\overline{Y})$ in the space $\overline {Y}$. Since the space $X$ is locally compact, there exists an open neighborhood $V$ of the point $x$ such that the set $\overline V$ is compact. Then $\psi(x,\overline{Y\cap V})\le \psi(x,\overline{Y})\le\omega$. Since the space $\overline {V\cap Y}$ is compact, the character $\chi(x,\overline {V\cap Y})$ is countable too. Since the set $Y$ is dense in the space $\overline{V\cap Y}$ and $x\not\in Y$, there exists a sequence $\{x_n\}$ of distinct points of the set $Y$  converging to $x$. Then the set $\{x_n\}\cup\{x\}$ is a countable infinite compact set, and hence an infinite closed subset of the space $X$ with countable interior, a contradiction.$\square$
Update. Moreover, we can improve Lemma to
Lemma 2. Each infinite Hausdorff space $X$ contains an infinite closed subspace with a countable interior.
Proof. Suppose the converse. Let $Y=\{y_n:n\in\omega\}$ be an arbitrary countable infinite subset of $X$. Then the set $\overline Y\backslash Y$ is uncountable. For each $n\in\omega$ we define by induction an open neighborhood $U_n$ of the point $y_n$ such that a set $Z_n=\overline Y\backslash (Y\cup\bigcup_{i\le n} U_i)$ is infinite and a point $z_n\in Z_n$ such that all the points $z_i$ are distinct and $z_i\not\in U_n$ for each $i<n$. If for each neighborhood $U$ of the point $y_0$ the set $\overline Y\backslash (Y\cup U)$ is finite, then for any countable infinite subset $Z$ of the set $\overline Y\backslash Y$ the set $Z\cup \{y_0\}$ is a countable compact set, a contradiction. Therefore we can choose an open neighborhood $U_0$ of the point $y_0$ such that the set $Z_0=\overline Y\backslash (Y\cup U)$ is infinite and a point $z_0\in Z_0$. If for each neighborhood $U$ of the point $y_1$ the set $Z_0\backslash U$ is finite, then for any countable infinite subset $Z$ of the set $Z_0$ the set $Z\cup \{y_1\}$ is a countable compact set, a contradiction. Therefore we can choose an open neighborhood $U_1\not\ni z_0$ of the point $y_1$ such that the set $Z_1=Z_0\backslash U_1$ is infinite and a point $z_1$ in $Z_1\backslash\{z_0\}$, and so forth. Then $\overline{\{z_n:n\in\omega\}}\subset\overline{Y}\backslash\bigcup U_n$ is a nowhere dense closed infinite subset of the space $\overline Y$, a contradiction.$\square$
Update 2. Moreover, we can prove 
Proposition. Each infinite Hausdorff space $X$ containing infinitely many non-isolated points contains also an infinite closed subset with empty interior.
Proof. Suppose the converse. Denote  the set of all non-isolated points of $X$ as $X’$. Let $Y=\{y_n:n\in\omega\}\subset X’$ be an arbitrary countable infinite subset. It is clear that $\overline Y\subset X’$.
We shall need the following 
Lemma 3. There are no infinite subset $Z$ of the set $X’$ and a point $x\in X$ such that for each neighborhood $U$ of the point $x$ the set $Z\backslash U$ is finite. 
Proof. Suppose the converse. Then the set $Z_0=Z\cup \{x\}$ is a compact set. Moreover, let 
$y\in Z_0\backslash \{x\}$ be an arbitrary point. There exist disjoint open sets $U\ni x$ and $V\ni y$. Hence $V\cap Z_0$ is a finite neighborhood of the point $y$. Therefore, the space $Z_0\backslash \{x\}$ is discrete. The set $Z_0$ is closed in $X$ as its compact subset. Since the set $Z_0$ is infinite, there exists a non-empty open subset $V$ of $X$ such that $V\subset Z_0$. Since the set $Z_0\backslash \{x\}$ is dense in $Z_0$, there exists a point $y\in V\cap (Z_0\backslash \{x\})$. Since the space $Z_0\backslash \{x\}$ is discrete, there exists a neighborhood $W$ of the point $y$ such that $W\cap (Z_0\backslash \{x\})$ is a singleton. But then a set $W\cap V\subset W\cap (Z_0\backslash \{x\})$ is a singleton too, a contradiction, because all points from the set $Z_0\backslash \{x\}\subset X’$ are non-isolated. $\square$
Now we shall consider two cases. 
$1.$ The set $\overline Y\backslash Y$ is infinite. For each $n\in\omega$ we define by induction an open neighborhood $U_n$ of the point $y_n$ such that a set $Z_n=\overline Y\backslash (Y\cup\bigcup_{i\le n} U_i)$ is infinite and a point $z_n\in Z_n$ such that all the points $z_i$ are distinct and $z_i\not\in U_n$ for each $i<n$. By Lemma 3, there exists an open neighborhood $U_0$ of the point $y_0$ such that the set $ Z_0=\overline Y\backslash (Y\cup U)$ is infinite. Choose an arbitrary point $z_0\in Z_0$. Again by Lemma 3, there exists an open neighborhood $U_1\not\ni z_0$ of the point $y_1$ such that the set $Z_1=Z_0\backslash U_1$ is infinite. Choose an arbitrary point $z_1\in Z_1\backslash\{z_0\}$, and so forth. Then $\overline{\{z_n:n\in\omega\}}\subset\overline{Y}\backslash\bigcup U_n$ is a nowhere dense closed infinite subset of the space $\overline Y$, a contradiction.
$2.$ The set $\overline Y\backslash Y$ is finite. Since $\overline Y\subset X’$, without loss of generality, we can assume that 
$\overline Y=Y$. For each $n\in\omega$ we define by induction an open neighborhood $U_n$ of the point $y_n$ such that a set $Z_n=Y\backslash \bigcup_{i\le n} U_i$ is infinite and a point $z_n\in Z_n$ such that all the points $z_i$ are distinct. By Lemma 3, there exists an open neighborhood $U_0$ of the point $y_0$ such that the set $Z_0=Y\backslash U_0$ is infinite. Choose an arbitrary point $z_0\in Z_0$. Again by Lemma 3, there exists an open neighborhood $U_1$ of the point $y_1$ such that the set $Z_1=Z_0\backslash U_1$ is infinite. Choose an arbitrary point $z_0\in Z_1\backslash\{z_0\}$, and so forth. Put $Z=\{z_i:i\in\omega\}$. Then $\overline Z\subset\overline Y=Y$. Let $y\in Y$. Then $y=y_n$ for some $n\in\omega$. Then the intersection $U_n\cap Z\subset U_n\cap (Z\backslash Z_n)$ is a finite set. Therefore $Z$ is a discrete (in particular, a closed) subset of the space $X$. Suppose that the set $Z$ has non-empty interior. Then there exists a non-empty open set $U$ such that $U\subset Z$. Then there exists 
a number $n$ such that $U_n\cap U$ is a nonempty open set. But $U_n\cap U\subset U_n\cap Z$ is a singleton, a contradiction, because all points from the set $Z$ are non-isolated. $\square$
