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**MadSoulz** We are in $\displaystyle (\mathbb{R}^2, d)$, where $\displaystyle d$ is the Euclidean metric. Let our region be the open ball centered at the point $\displaystyle x=(-2,0)$ with radius 1, denoted by $\displaystyle B(x;1)$.

My attempt: Let $\displaystyle F_n= B(x;1-\frac{1}{n})$ where $\displaystyle n\in \mathbb{N}$.

It's obvious to me (not sure if I must prove) that $\displaystyle \{ F_n \}$ is an open cover for $\displaystyle B(x;1)$ since $\displaystyle B(x;1) \subset \bigcup_{n\in\mathbb{N}} F_n$

Suppose there exists a finite subcover of $\displaystyle \{ F_n \}$, say $\displaystyle \{F_{n_{i}}\}$. I must find an element in $\displaystyle t \in B(x;1)$ such that $\displaystyle t \notin \{F_{n_{i}}\}$ to reach a contradiction, right?

Since $\displaystyle \{F_{n_{i}}\}$ is a finite subcover we have that $\displaystyle B(x;1) \subset \bigcup_{n=1}^{n_i} F_{n_i}$. So every point in $\displaystyle B(x;1)$ is also in $\displaystyle B(x; 1-\frac{1}{n_i})$. Consider $\displaystyle n_i + 1$. It's obvious that $\displaystyle 1-\frac{1}{n_i} < 1 - \frac{1}{n_i+1} < 1$, so

$\displaystyle B(x; 1-\frac{1}{n_i + 1}) \not\subset B(x;1-\frac{1}{n_i})$since the radius is bigger, but

$\displaystyle B(x; 1-\frac{1}{n_i + 1}) \subset B(x;1)$ since the radius is still less than 1. We have reached a contradiction.