Originally Posted by

**aharonidan**

b. Space $\displaystyle X$. every infinite set $\displaystyle S \subseteq X $has an accumulation point and $\displaystyle X$ is not compact.

Consider $\displaystyle X=[0,1]^\Omega$ where $\displaystyle \#(\Omega)>\aleph_0$ with the product topology and define

$\displaystyle \mathcal{X}=\left\{(x_\omega)\in X:x_\omega=0\text{ for all but countably many }\omega\in\Omega\right\}$

Let $\displaystyle Y$ be an infinite subset of $\displaystyle \mathcal{X}$. That space does the trick. Try proving it.