Prove $\overline{x+A} = x+\bar{A}$ and $\alpha\bar{A} = \overline{\alpha A}$ If $A$ is a subset of $(V,\parallel.\parallel)$, then let $\bar{A}$ denote its closure. Show that if $x\in V$ and $\alpha \in \mathbb{R}$, then $\overline{x+A} = x+\bar{A}$ and $\alpha\bar{A} = \overline{\alpha A}$.
Honestly, I'm pretty stumped as to where to start, any help would be appreciated.

Another approach:  $y \in \overline{x + A}$ iff there's a sequence $\{x + a_n\}_{n=1}^{\infty}$ so that $x + a_n \to y$ as $n \to \infty$, which happens iff $a_n \to y - x$ as $n \to \infty$; so $(y-x) \in \overline{A}$, and so $y \in x + \overline{A}$.
A similar method shows that $\alpha \overline{A} = \overline{\alpha A}$.
