Suppose $A\subseteq \mathbb{R}$ is closed and non-empty. Prove that $2A$ is closed., where $2A=\{2a: a\in A\}$
Let $(x_n)_{n\in\Bbb N}$ be a convergent sequence of elements of $2A$, and let $x$ be its limit. For each $n\in\Bbb N$, let $a_n\in A$ be such that $2a_n=x_n$. Then $(\forall n\in\Bbb N):a_n=\frac{x_n}2$, and therefore$$\frac x2=\lim_{n\to\infty}\frac{x_n}2=\lim_{n\to\infty}a_n.$$But $\lim_{n\to\infty}a_n\in A$, since $A$ is closed. In other words, $\frac x2\in A$. Therefore, $x\in2A$. Since the limit of every convergent sequence of elements of $2A$ belongs to $2A$, $2A$ is a closed set.