The answer is no. Suppose $X$ is such a space. Since closed subsets of compact spaces are themselves compact, every closed proper subset of $X$ is compact and therefore finite.
Now let $S$ be an arbitrary infinite subset of $X$ and let $\mathcal C$ be an open cover of $S$. Suppose $G$ is any element of $\mathcal C$. Then $S\setminus G$ is finite because it is the intersection of $S$ and the closed (and therefore finite) set $X\setminus G$. The finite set $S\setminus G$ is easily covered by a finite subcover $\mathcal C'\subset \mathcal C$, so $\mathcal C'\cup\{G\}$ is a finite subcover of $S$. This shows that the infinite set $S$ is compact, a contradiction.
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Let $X$ be an infinite compact space, and choose an injective sequence in $X$ that misses at least two points in $X$. Then the union of a convergent subsequence and its limit is a proper, infinite, compact subset.
The answer is no. Suppose $X$ is such a space. Since closed subsets of compact spaces are themselves compact, every closed proper subset of $X$ is compact and therefore finite.
Now let $S$ be an arbitrary infinite subset of $X$ and let $\mathcal C$ be an open cover of $S$. Suppose $G$ is any element of $\mathcal C$. Then $S\setminus G$ is finite because it is the intersection of $S$ and the closed (and therefore finite) set $X\setminus G$. The finite set $S\setminus G$ is easily covered by a finite subcover $\mathcal C'\subset \mathcal C$, so $\mathcal C'\cup\{G\}$ is a finite subcover of $S$. This shows that the infinite set $S$ is compact, a contradiction.