Proving that $\|A\|_{\infty}$ the largest row sum of absolute value of matrix $A$

I assume that the author tries to derive the matrix norms $\|A\|_1$ and $\|A\|_\infty$ induced by vector norms $\|x\|_1$ and $\|x\|_\infty$.

Let's take, for example, $\|\cdot\|_\infty$. We write

$$\|Ax\|_\infty=\sup_i \left|\sum_j A_{ij}x_j\right|\le \sup_i \sum_j |A_{ij}||x_j|\le \sup_j |x_j| \sup_i \sum_j |A_{ij}|.$$ Hence, a good candidate for $\|A\|_\infty$ is $\sup_i \sum_j |A_{ij}|$. We need to prove that this boundary is indeed achieved; it is true, since we can take $i$ where that supremum is achieved and impose $x_j=\mathrm{sign}(A_{ij})$. With such $x$ all our inequalities degenerate to equalities and we conclude that $\|A\|_\infty = \sup_i \sum_j |A_{ij}|$ is a matrix norm induced by $\|x\|_\infty = \sup_j |x_j|$.

The case $\|\cdot\|_1$ is done likewise.

I will give you some guidelines then. Suppose an $m$-by-$n$ matrix $A = (a_{ij})$ is given. For any $n$-dimensional vector $x$,

\begin{align*} \|Ax\|_1 & = \sum_{i=1}^m \left|\sum_{j=1}^n a_{ij}x_j\right| \\ & \le \sum_{i=1}^m \sum_{j=1}^n \left|a_{ij}x_j\right| \\ & = \sum_{j=1}^n \left|x_j\right| \sum_{i=1}^m \left|a_{ij}\right| \\ & = \sum_{j=1}^n \left|x_j\right| A_j \end{align*} where I define $A_j = \sum_{i=1}^m \left|a_{ij}\right|$. If $J = \operatorname{argmax}_j A_j$, i.e., $A_J$ is a maximum among all $A_j$'s, then $$ \|Ax\|_1 \le A_J \sum_{j=1}^n |x_j| = A_J \|x\|_1. $$ This shows that $\|A\|_1 \le A_J$. Next, you show that there exists $x$ such that $\|Ax\|_1 = A_J\|x\|_1$. This is quite simple as you can choose $x_i = 0$ for all $i \ne J$, and $x_J = 1$ or $-1$.

For the case of $\|\cdot\|_\infty$, the situation is quite similar. (Choosing $x$ in the last step might be a bit tricky as you need to pick the right signs.) I will leave that to you.