Be $\mathbb{R}$ with Sorgenfrey's topology where the basis elements are of the form $[a,b)$. If $A\subseteq\mathbb R$ is compact, then $A$ is countable.
I tried to find a set $A$ not countable which has a cover that can not be reduced to a finite one, but I could not find it. Is it possible to do it another way?
A simple example: $[0,1]$ is not compact in the Sorgenfrey topology because the open cover $U_n = [0, 1-\frac{1}{n}), n \ge 2$ together with $[1,2)$ is an open cover of $[0,1]$ that has no finite subcover: suppose it had: let $k$ be the largest of the $n$ used in the first type of open set; then $1-\frac{1}{k+1}$ is not covered by any of the sets in the cover (as all right hand points of the $U_n$ in the cover are smaller by construction, and $[1,2)$ is only there to cover $1$).
HINT: This answer shows that if $A$ is uncountable, there is an uncountable $A_0\subseteq A$ such that if $V$ is an open interval in $\Bbb R$, then either $V\cap A_0=\varnothing$, or $V\cap A_0$ is uncountable. Suppose that for each $a\in A_0$ there is an $x_a<a$ such that $(x_a,a)\cap A_0=\varnothing$. Then $\{(x_a,a):a\in A_0\}$ is an uncountable collection of pairwise disjoint non-empty open intervals in $\Bbb R$, which is impossible: each of these intervals must contain a distinct rational number, but there are only countably many rational numbers. Thus, there must be an $a_0\in A_0$ such that $(x,a_0)\cap A_0$ is uncountable for each $x<a_0$. Now consider the open cover
$$\{(\leftarrow,x):x<a_0\}\cup\{[a_0,\to)\}$$
of $A$.
Hint. A set $A\subseteq\mathbb R$ is compact in the Sorgenfrey topology if and only if it satisfies the two conditions:
(1) $A$ is compact in the usual topology of $\mathbb R$;
(2) $A$ is a reverse well-ordered set, i.e., every nonempty subset of $A$ has a greatest element.
Any set $A\subseteq\mathbb R$ satisfying (2) is countable.
