KC implies US

Statement
Any KC-space is a US-space.

KC-space
A topological space is termed a KC-space if every compact subset of it is closed.

US-space
A topological space is termed a US-space if every convergent sequence has a unique limit.

Proof
Given: A KC-space $$X$$.

To prove: If $$\{ x_n \}_{n \in \mathbb{N}}$$ is a sequence with limit $$x$$ and limit $$y$$, then $$x = y$$.

Proof: Clearly, $$x_n$$ cannot be eventually constant at both $$x$$ and $$y$$. Let us say that it is not eventually constant at $$y$$. Throw out all the $$x_n$$s that are equal to $$y$$. This new sequence of $$x_n$$s converges to both $$x$$ and $$y$$. Let $$A$$ be the union of $$\{ x \}$$ and the $$x_n$$s in this new sequence.


 * 1) $$A$$ is compact: Any open cover of $$A$$ contains one open set containing $$A$$, so it contains all but finitely many of the $$x_n$$s. Thus, that along with finitely many other open subsets covers $$A$$. Thus, every open cover has a finite subcover.
 * 2) $$A$$ is closed: This follows from the previous step, and the assumption that $$X$$ is a KC-space.
 * 3) There is an open subset containing $$y$$ that does not contain any of the $$x_n$$s: This follows from the fact that $$A$$ is closed, and $$y \notin A$$.

The last step contradicts our assumption that the $$x_n$$s converge to $$y$$.