Show that the subset $A=\{(x_1,...,x_n)\in\Bbb R^n |−1≤x_1 ≤x_2 ≤···≤x_n ≤1\}$ is compact.
A is contain in an open cover as it is contained in $\Bbb R^n$. Therefore there exists a finite sub cover which is the union of the open balls centred at $x_i$ for $i \in I$ with radius $>2$ implying compactedness. Am I on the right track or where have i gone wrong?
You should proceed using Heine-Borel theorem. The set you presented is clearly bounded, and it is closed, therefore, it is compact.
To check that it is closed, simply consider any point $y\notin A$. Then, see that for a sufficient small neighborhood of $y$, $V_y$, then $V_y \subset \mathbb{R}^n\setminus A$. This will follow simply assuming that one of the inequalities defining $A$ does not hold, and seeing that it will not hold for points near $y$.