**Problem.** Assume that a compact space $X$ can be written as the union $X=K\cup D$ of a compact metrizable subspace $K$ and a discrete subspace $D$.
Does $D$ contain a non-trivial convergent sequence in $X$?


As shown by Ilya Bogdanov ([Does every compact countable space contain a non-trivial convergent sequence?](https://mathoverflow.net/questions/254211/does-every-compact-countable-space-contain-a-non-trivial-convergent-sequence)) the answer is affirmative if $K$ is at most countable.



Convergent sequences in compact spaces
Oh, sorry! I wrote this question and after some thinking found a (relatively simple) answer.


Consider the set $\mathcal P$ of pairs $(A,I)$ where $A$ is a non-empty closed subset of the compact metrizable space $K$ and $I$ is an infinite subset of $D$ such that every open neighborhood $U$ of $A$ in $X=K\cup D$ contains all but finitely many points of the set $I$. The set $\mathcal P$ is endowed with the partial order: $(A,I)\le (B,J)$ if $A\subset B$ and $I\setminus J$ is finite. Let $(A\_\alpha,I\_\alpha)\_{\alpha<\kappa}$ be a maximal descreasing transfinite sequence of elements of $\mathcal P$ such that for any ordinals $\alpha<\beta<\kappa$ the set $A\_\beta$ is a proper subset of $A\_\alpha$. The second countability (in fact, the hereditary Lindelofness) of $K$ guarantees that the length $\kappa$ of such sequence is countable. Moeover, the ordinal $\kappa$ is successor. In the opposite case we can extend the transfinite sequence $(A\_\alpha,I\_\alpha)\_{\alpha<\kappa}$ letting $A\_\kappa=\bigcap\_{\alpha<\kappa}A\_\alpha$ and $I\_\kappa$ be any pseudointersection of the countable tower $(I\_\alpha)\_{\alpha<\kappa}$. So, $\kappa=\lambda+1$ for some ordinal $\lambda$. We claim that $A\_\lambda$ is a singleton, and hence $I\_\lambda$ is a required non-trivial convergent sequence. If $A\_\lambda$ is not a singleton, we can write it as the union $A\_\lambda=B\cup C$ of two proper closed subsets of $A\_\lambda$. By the maximality of the transfinite sequence $(A\_\alpha,I\_\alpha)\_{\alpha\le\lambda}$, the set $B$ has an open neighborhood $U\subset X$ such that $J=I\_\lambda\setminus U$ is infinite. Using the maximality once more, we can find an open neighborhood $V\subset X$ of $C$ such that $J\setminus V$ is infinite. Then $U\cup V$ is an open neighborhood of $A\_\lambda$ in $X$ such that $I\_\lambda\setminus (U\cup V)=J\setminus V$ is infinite, which contradicts the inclusion $(A\_\lambda,I\_\lambda)\in\mathcal P$.


