Verify proof: compact iff finite with discrete metric in $\mathbb{R}^n$ Consider $\mathbb{R}^n$ with the discrete metric $d$, i.e., 
$$d(x, y) = \begin{cases} 0 & \text{ if } x = y \\ 1 &\text{ if } x \neq y \end{cases}$$
Using the open cover definition of compactness, show that $X \subseteq \mathbb{R}^n$ is compact if and only if $X$ is a finite set.
Could someone verify whether my proof is correct or not?
First for the ($\implies$) direction. Assume that $X$ is compact, then every open cover of $X$ has a finite subcover. Consider the particular open cover $U_{\varepsilon} = \{B_{\varepsilon}(x) : x \in X\}$ for $0<\varepsilon<1$. Then there exists $x_1, x_2, \cdots, x_n$ such that $X \subseteq B_{\varepsilon}(x_1) \cup \cdots \cup B_{\varepsilon}(x_n)$. But $B_{\varepsilon}(x_i) = \{x_i\}$ for $i=1, \cdots, n$ because we have the discrete metric. So $X \subseteq \{x_1, \cdots, x_n\}$ and hence $X$ must be finite.
For the ($\impliedby$ ) direction. Assume $X$ is finite, i.e., $X = \{x_1, \cdots, x_n\}$. Now let $U = \{U_{\lambda} : \lambda \in \Lambda\}$ be any open cover for $X$. This means that $x_i \in U_{\lambda_i}$ for some $\lambda_i \in \Lambda$ for all $i = 1, \cdots, n$. So $X \subseteq U_{\lambda_1} \cup \cdots \cup U_{\lambda_n}$ and so we have produced a finite subcover. 

In the second part, since $X $ is finite,  you can take, for each $x\in X$, one open set $U_x$, containing $x$, and you will have a finite subcover. 
This is a pretty easy result; and I'd say your proof is fine...
The point I think is that when the points are open sets,  you obviously can't have infinitely many of them, or else you have an infinite cover with no finite subcover. ..
