Here is a way to show this when you consider the field of real numbers and this for all norms.
Consider $E$ a finite dimensional vector space over $\mathbb{R}$ and a basis $(e_1, ..., e_n)$ of $E$. It means that for all $x\in E$ you have $x = \sum_{i=1}^{n} x_{i}e_i $ where the $x_i$ are real numbers.
Now we need to endow E with a norm, first recall that all norms are equivalents in a finite dimensional normed space it will be usefull at some points.
The norm we will consider is $\lVert x\rVert_{\infty} = max_{1\leq i\leq n}\lvert x_i\rvert $. Thus $(E,\lVert\rVert_{\infty})$ is a finite dimensional normed space.
Now consider a Cauchy sequence in $E$, namely $(x^{p})_{p\in\mathbb{N}}$, it means that
$\forall\epsilon>0,\exists N\in\mathbb{N} : n,m\geq N\implies\lvert x_i^{n} -x_i^{m}\rvert\leq\lVert x^{n} - x^{m}\rVert_{\infty} = max_{1\leq i \leq n}\lvert x_{i}^{n} - x_{i}^{m}\rvert\leq\epsilon$
However for $i\in\{1,...,n\}$ we have that $(x_i^{p})$ is a sequence of real numbers and we have shown that it is Cauchy !
Thus for all $i\in\{1,..n\}$ there exists $x_i\in\mathbb{R}$ such that $\forall\epsilon>0, \exists N_{i}\in\mathbb{N} : p\geq N_i\implies \lvert x_i^{p} -x_i\rvert\leq\epsilon$ (since $\mathbb{R}$ is complete)
Now, a candidate we should consider as a limit of $(x^p)_{p\in\mathbb{N}}$ is the vector whose coordinate corresponds to the $x_i$'s that is $x = \sum_{i=1}^{n}x_ie_i$.
Fix $\epsilon>0$ and $N = max_{1\leq i\leq n}(N_i)$, then we have that
$p\geq N\implies max_{1\leq i\leq n}\lvert x_i^p - x_i\rvert = \lVert x^p - x\rVert_{\infty}\leq\epsilon$ which shows the convergence of $(x^p)_{p\in\mathbb{N}}$ to $x\in E$.
Since $E$ is a finite dimensional normed space, all norms on this space are equivalents which means that the convergence is not altered by a change of norm (this is already explained in the comments I think)
I hope this is clear !
Best Answer
To show that $A_n \to A$ in $L(E,F)$, we need to show that $||A_n - A|| \to 0$, which is equivalent to saying that for any $\epsilon > 0$, there exists $N$ such that if $n > N$ , we have $||A_n - A|| < \epsilon$.
However, $||A_n - A|| < \epsilon$ is the same as $||(A_n - A)x|| < \epsilon ||x||$ for all $x \in E$ , which is the same as $||A_n x - Ax|| < \epsilon ||x||$ for all $x \in E$, which is the statement you have.
In other words, the last statement is the assertion that $||A_n - A|| < \epsilon$, which was to be shown for showing $A_n \to A$. This equivalence was omitted in the text since it nearly follows by definition.