Proof: Let $\mathscr{U}$ := {$U_n$}$_{n \geq 1}$ = {$(n-1,n+1)$}$_{n \geq 1}$ be an open cover of $\mathbb{R}$. Consider finitely many of them, say: $U_{n_1} \ldots U_{n_k}$. Setting $N := \max(n_1 \ldots n_k)$, we get that: $U_{n_i} \subset U_N \implies \bigcup_{i=1}^{k} U_{n_i} = U_N = (N-1,N+1) \subsetneq \mathbb{R}$.
Hence, $\mathscr{U}$ cannot be reduced to a finite subcover of $\mathbb{R}$ so that it is not compact. QED.
How did I do? Thanks.
Best Answer
As pointed out in the comments by Fred and Levent, $\mathcal U$ is not an open cover of $\mathbb R$ since $\bigcup_{n\geqslant 1}(n-1,n+1)=(-1,+\infty)$. However, this can be fixed easily: define instead $U_n:=(-n,n)$. Then $\bigcup_{n\geqslant 1}U_n=\mathbb R$ and a finite subcover is contained in an interval of the form $(-R,R)$.