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**Plato** I assume that we are in a metric space.

Here are traditional proofs for both properties.

Suppose that $\displaystyle x$ is a limit point of $\displaystyle K$ but $\displaystyle x\notin K$.

$\displaystyle \left( {\forall y \in K} \right)$ this is true $\displaystyle r_y = \frac{{d(x,y)}}{4} > 0$.

The collection $\displaystyle \left\{ {B(y;r_y )} \right\}_{y \in K} $ covers $\displaystyle K$.

So finite subcollection $\displaystyle K \subset \bigcup\limits_{j = 1}^n {B(y_j ;r_{y_j } )}$ also covers $\displaystyle K$.

But note that $\displaystyle x \in \bigcap\limits_{j = 1}^n {B(x ;r_{y_j } )}$. **That is a open set that contains $\displaystyle x$ and no other point of $\displaystyle K$. **Contradiction.

For bounded, there is finite collection $\displaystyle \bigcup\limits_{j = 1}^n {B(y_j ;1)} $ covering $\displaystyle K$.

Let $\displaystyle M = \max \left\{ {d(y_k ,y_j )} \right\} + 2$. It is easy to show that $\displaystyle M$ is a bound for$\displaystyle K$.