Let $A \subset \mathbb{R}$. Show that $A$ is sequentially compact if and only if $A$ is compact.
I have looked for other explanations of this, but I can't find one that is for $\mathbb{R}$ specifically, and I don't yet fully understand metric spaces.
Suppose $A$ is compact, and consider a sequence $(x_n) \subset A$. Suppose this did not have a convergent subsequence, then $\forall x\in A, \exists \epsilon_x > 0$ such that $$ U_x := (x-\epsilon_x, x+\epsilon_x) $$ contains atmost one point of the sequence (why?). Now the collection $\{U_x : x\in A\}$ forms an open cover for $A$ and so has a finite subcover. Conclude from this that the set $\{x_n\}$ is a finite set - a contradiction.
Perhaps you can try the reverse direction, and let me know if you have trouble and I could complete the proof. Hope this helps.