Show Euclidean plane with usual topology is not compact

Question

I need to show an open cover with no finite subcover.

My try

I assume Euclidean plane with usual topology deals with open sets/intervals.
So

Let A=(a,b). Let $G_n$:=(a-(1/n),b) for each
n$\in $N then A =$\bigcup_{r=1}^{\infty}G_n$. So
G is an open cover for A
. Let {$G_{n_1},G_{n_2}$,…,$G_{n_k}$ } be
a subcollection of $G_n$ .Take
K=max{$n_1,n_2$,…,$n_k$ }
then $\bigcup_{r=1}^{K}$ $G_{n_r}$= $G_n$ =(a-($1$/K),b)

Since $a-(1$\K)$\in$ A$\notin$ $G_K$ ,no finite subcover can be found.

Therefore A is not compact.

I hope I solved it.

Any feedback would be appreciated.

Answer 1

Consider a family of open sets $$U_n =\{(-n,n): n \in \mathbb{N}\}$$

$$\mathbb{R}=\cup_{n \in \mathbb{N}}(-n,n)$$

But this open cover has no finite subcover.