Suppose $F$ is a non-Archimedean local field, thus $F$ is the field of fractions of a discrete valuation ring $\mathfrak{o} $. Let $\mathfrak{p} $ be the maximal ideal of $\mathfrak{o}$, if $\mathfrak{w}$ is the prime element of $F$,then $x\in \,F$ can be write as $x=u\mathfrak{w}^n$, which $u\in \,\mathfrak{o} ^{\times},n\in \mathbb{Z}$.
The $F$ carries an absolute value $$\left\| x \right\| =q^{-n},\left\| 0 \right\| =0, q>0 $$
Which give a metric on $F$.
My questions are: why
$$\mathfrak{p} ^n=\left\{ x\in \,\,F:\left\| x \right\| \leqslant q^{-n} \right\} ,n\in \mathbb{Z} \,\,\,\,\,\,\,\,\,\,\,(1)$$
are open subsets of $F$?
I'm not sure I fully understand your question, partly because there seem to be some undefined symbols ($q$, $a$).
Is your question why a set of the form $\left\{x\in F\mid \lVert x\rVert\leq q^{-n}\right\}$ for some $n$ is open? Intuitively, this is because there are gaps between the possible values of the norm. Assuming $q>1$, you find $$\left\{x\in F\mid \lVert x\rVert\leq q^{-n}\right\}=\left\{x\in F\mid \lVert x\rVert< \frac{q^{-(n-1)}+q^{-n}}{2}\right\}.$$ which is an open set.
The reason for this equality is as follows. We have $q^{-(n-1)}>q^{-n}$, so any $x$ in the set on the left is also in the set on the right. On the other hand, if $x$ is not in the set on the left, then $q^{-n}<\lVert x\rVert$. But then $\lVert x\rVert$ is at least $q^{-(n-1)}$, so $x$ is also not in the set on the right.
Note that $\frac{q^{-(n-1)}+q^{-n}}{2}$ is more or less arbitrary. We could have taken any value $a$ with $q^{-n}<a<q^{-(n-1)}$.