Definition of $L_\infty$ norm Where does the definition of the $L_\infty$ norm come from?
$$\|x\|_\infty=\max \{|x_1|,\dots,|x_k|\}$$
 A: Here's the way I like to think about it (which is not too rigorous but can be made rigorous). We have, more generally, the $L^p$ norms ($1\le p < \infty$):
$$(\|x\|_p)^p:=\sum_{i=1}^n |x_i|^p.$$
The Euclidean norm is a special case of this (take $p = 2$); the taxicab norm is also a special case (take $p=1$). Suppose $|x_i|\ge |x_j|$ for all $1\le j\le n$. What happens if $p$ gets really large? Well we would see that the $|x_i|$ term would dominate the sum and asymptotically, all of the others would be inconsequential (due to the function being very convex). So what we could say is that for large $p$,
$$(\|x\|_p)^p \approx |x_i|^p.$$
Taking a $p$th root of both sides gives us
$$\|x\|_p \approx |x_i|.$$
However note that we said that $|x_i|$ was the largest of the components of our vector. Thus, informally, we would say that $\|x\|_{\infty} = |x_i| = \max \{|x_1|,\ldots,|x_n|\}$.
A: The $p$ norm of a vector is defined as such:
$\|x\|_p = (\sum_{i=1}^{n}|x_i|^p)^\frac{1}{p}$.
Notice that when $p=2$ this is the simple euclidean norm.
You asked about the infinity norm.
When $p$ tends to infinity, we can see that:
$$\lim_ {p \to \infty} \|x\|_p = \lim_ {p \to \infty} (\sum_{i=1}^{n}|x_i|^p)^\frac{1}{p}$$
Convince yourself that if $a>b>0$ then:
$\lim_{p \to \infty} a^p+b^p=\lim_{p \to \infty} a^p$ 
Combine the two statements to reach the desired results. 
A: General Approach
As shown in this answer,
$$
\lim_{p\to\infty}\left(\int_A|f(x)|^p\,\mathrm{d}x\right)^{1/p}=\sup_{x\in A}|f(x)|\tag{1}
$$
Using a discete measure, $(1)$ shows that
$$
\lim_{p\to\infty}\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}=\max_{1\le k\le n}|x_k|\tag{2}
$$
Since
$$
\|x_k\|_p=\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}\tag{3}
$$
if we take the limit of $(3)$ and compare with $(2)$, we get
$$
\|x_k\|_\infty=\max_{1\le k\le n}|x_k|\tag{4}
$$

Discrete Approach
The discrete $\ell^p$ norm is defined as
$$
\|x_k\|_p=\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}\tag{5}
$$
suppose that $m=\max\limits_{1\le k\le n}|x_k|$, and that $q$ of the $|x_k|$ equal $m$ (that is, $1\le q\le n$). For the $|x_k|\lt m$, we have $\lim\limits_{p\to\infty}\left|\frac{x_k}{m}\right|^p=0$. Therefore,
$$
\begin{align}
\lim_{p\to\infty}\|x_k\|_p
&=\lim_{p\to\infty}\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}\\
&=m\lim_{p\to\infty}\left(\sum_{k=1}^n\left|\frac{x_k}{m}\right|^p\right)^{1/p}\\[4pt]
&=m\lim_{p\to\infty}q^{1/p}\\[12pt]
&=m\tag{6}
\end{align}
$$
Taking the limit of $(6)$, we get that
$$
\|x_k\|_\infty=\max_{1\le k\le n}|x_k|\tag{7}
$$
A: The following adds nothing to the above proofs other than a slightly geometric flavour. It is too long for a comment, so I am including it as an answer.
I am taking the space to be $\mathbb{C}^n$.
It is not hard to show (see https://math.stackexchange.com/a/424335/27978 for example) that $p \mapsto \|x\|_p$ is non-increasing,
$\lim_{p \to \infty} \|x\|_p = \|x\|_\infty$, and $\|x\|_\infty \le \|x\|_p \le \|x\|_1 \le \sqrt{n} \|x\|_\infty$, hence  the concepts of interior and boundedness are independent of $p$.
The above shows that $B_\infty(0,1) = \cup_{ p \ge 1 } B_p(0,1)$.
Since each $\|\cdot \|_p$ is a norm, we see that each $B_p(0,1)$ is bounded (uniformly by comment above), balanced, convex and contains $0$ in its interior. Hence $\cup_{ p \ge 1 } B_p(0,1)$ is bounded, balanced, convex and contains $0$ in its interior, and so is the unit ball of some norm, in this case, $\|\cdot \|_\infty$.
The point here is that there is a natural motivation in terms of the unit balls which 'converge' (in the above sense) to the $\|\cdot \|_\infty$ unit ball.
