Suppose $X$ is a metric space, $A\subset X$ and $A\subset A'$(the set of limit points of $A$). Prove that $A$ is sequentially compact if and only if every continuous function on $A$ is uniformly continuous.
The "$\Rightarrow$" is easy and is already on MSE. For the "$\Leftarrow$", I tried to prove $A$ is closed as follows:
If $A=A'$, then A is closed, otherwise we can choose $x\in A'\backslash A$, and there exisets a sequence $x_n\to x$, $x_n\in A$ and are different from each other. I had trouble to define an approciate continuous function and don't know how to go on.
Appreciate any help!