The title is pretty almost self explanatory, but I will leave my precise question below.
Question. Consider the normed space $(\Bbb R^n,\|\cdot\|_\infty)$, where the infinite norm is the usual one. Show that the unit sphere defined by $$ S = \{\alpha \in \Bbb R^n \colon \|\alpha\|_\infty = 1\}$$ is closed and bounded.
My attempt. $S$ is obviously bounded (each element of $S$ satisfies the property $\|\alpha\|_\infty = 1$ which implies that each element of $S$ also satisfies the property $\|\alpha\|_\infty \leqslant 1).$
Now, I was trying to prove $S$ is closed (directly) but I am having some troubles. To see $S$ is closed, it suffices to prove that $\overline{S} \subset S.$ Let $\alpha \in \overline{S}$ be an arbitrary point. Then, there exists a sequence $(\alpha_k)_{k \in \Bbb N}$ s.t. $\alpha_k \rightarrow \alpha$ as $k \rightarrow \infty.$ By definition of a convergent sequence,
$$\forall \epsilon > 0, \exists N = N(\epsilon) \colon \forall n \in \Bbb N, n > N \Rightarrow \|\alpha_n-\alpha\|_\infty < \epsilon. $$ But $$\|\alpha\|_\infty = \|(\alpha-\alpha_n) + \alpha_n\|_\infty \leqslant \|\alpha-\alpha_k\|_\infty + \|\alpha_n\|_\infty < \epsilon +1. $$ From here I don't know how to proceed. Any help would be apreciated.
The essential property you are missing is that the norm is continuous (basically by definition). This means that your proof gets even simpler: Take $(a_n)_n$ so $a_n\to a$, with $a_n\in S$. Then, as the norm is continuous and therefore commutes with limits, $1=\lim_{n\to\infty}||a_n||_\infty =||\lim_{n\to\infty}a_n||_\infty=||a||_\infty $.
Another way to see it is that since $||\cdot||_\infty:\mathbb{K}^n\to\mathbb{R}_{\geq 0}$ is continuous, the preimage of closed sets are closed. And $S$ is exactly the preimage of $\lbrace 1\rbrace$.