Heine-Borel Theorem ($\mathbb{R}^k$) (in ZF) Heine-Borel Theorem;
If $E \subset \mathbb{R}^k$, then $E$ is compact iff $E$ is closed and bounded.
I have proved 'closed and bounded⇒compact' and 'compact⇒bounded'.
(There exists $r\in \mathbb{R}$ such that for every $x\in E$, $|x|<r$)
The proof in Rudin PMA p.40 uses 'countable axiom of choice'
I have googled it and found some proofs, but they all used some weaker form of AC.
Please help me how to show that $compact⇒closed$ in ZF..
 A: We will prove that compact implies closed by contraposition.
Suppose that $E$ is not closed. Then there is some $x\notin E$ such that every neighbourhood of $x$ has a nonempty intersection with $E$. In particular, the collection $E_n:=E\setminus \overline {B(x,1/n)}$ (where $\overline {B(x,1/n)}$ is the closed ball centered at $x$ with radius $1/n$) is an infinite, nondecreasing open cover of $E$ (because for any $p\in E$ and $n>1/d(x,p)$ we have $p\in E_n$).
It is enough to show that $E_n$ does not stabilize. But if it did, then we would have for some $N<\omega$ that $E_N=E$, so $B(x,1/N)$ would be disjoint from $E$, so $x$ would not belong to the closure of $E$, so we're done.
This argument should work in an arbitrary Hausdorff space, though without countable character the cover will not be a sequence, but a directed set.
A: It is a bit late, but since Ink's answers may by written without the axiom of choice, I think that my answer is also pertinent.
We fix a Hausdorff space $X$, $A\subset X$ compact, $x\in X\setminus A$ and let $\mathcal{C}=\{(C,C'): C,C'\subset X$ are disjoint open sets such that $x\in C'\}$ and notice that $\text{dom}(\mathcal{C})=\{C:\exists C'$ s.t. $(C,C')\in \mathcal{C}\}$ is an open covering for $A$, because $X$ is a Hausdorff space.
Now, the compactness of $A$ implies the existence of $\{C_0,\dotso,C_n\}\subset\text{dom}(\mathcal{C})$ such that $A\subset C_0\cup\dotso\cup C_n$, and the correspondent open sets $C_0',\dotso,C_n'$ satisfy $x\in\bigcap_{j\leq n}C_j'\subset X\setminus A$.
This shows that $X\setminus A$ is open, as desired.
A: Any compact subset of a Hausdorff space is closed.
Proof:  Let $A \subset X$ be compact.  We show that $X -A$ is open.  Let  $x \in X - A$. Since $X$ is Hausdorff, for every $a \in A$, there are disjoint open sets $U_a$ and $V_a$ such that $x \in U_a$ and $a \in V_a.$ So $\{V_a\}_{a \in A}$ is an open cover which has  a finite subcover $\{V_{a_1},...,V_{a_n}\}$.  Let $V =\bigcup\limits_{i=1}^n V_{a_i}$ and $U = \bigcap\limits_{i=1}^n U_{a_i}$. Clearly, $U \cap V = \emptyset$, so $x \in U \subset X -A$.  Hence, $X - A$ is open.
