Show that the set $[0,1]×[0,1]$ is compact in $\mathbb{R}^2$ to standard metric

Question

I have to show that the set $[0,1]×[0,1]$ is compact in $\mathbb{R}^2$ with respect to the standard metric.

I have to show this using only the definition of compactness. The definition I am given is: A set is compact if we have an open cover, we get a finite subcover.

Answer 1

Heine-Borel Theorem states that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and your subset is certainly closed and bounded

Answer 2

Hints:

Every closed interval in $\mathbb{R}$ is compact.

The product of finitely many compact spaces is compact.

Answer 3

Let $\{U_i\}_{i=1}^\infty $ a cover of open set of $[0,1]\times [0,1]$. Since $U_i$ is open, there are open cubes $(a_j^i,b_j^i)\times (a_j^i,b_j^i)$ s.t. $$U_i=\bigcup_{j=1}^\infty (a_j^i,b_j^i)\times (a_j^i,b_j^i).$$

Therefore $(a_j^i,b_j^i)_{i,j=1}^{\infty }$ is an open cover of $[0,1]$. Therefore, there is a subcover (denoted $\{(a_j^i,b_j^i)\}_{\substack{i=1,...,n\\ j=1,...,m}}$) of $[0,1]$. Therefore $\{U_i\}_{i=1}^n$ is a finite subcover of $[0,1]\times [0,1]$.