A consequence of "every norm is equivalent to the sup-norm" in a finite dimensional normed vector space

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So, I am reading the following proposition in Neukirch's Algebraic Number Theory:

Proposition. Let $K$ be a complete field with respect to the valuation $|\:\: |$ and let V be an $n$-dimensional normed vector space over $K$. Then, for any basis $v_{1},...,v_{n}$ of $V$, the maximum norm

\begin{equation} ||(x_{1}v_{1}+\dots x_{n}v_{n}||=\max\{|x_{1}|,\dots , |x_{n}|\} \end{equation}

is equivalent to the given norm on $V$. In particular, $V$ is complete and the isomorphism

\begin{equation} K^{n}\to V,\qquad\qquad (x_{1},\:\dots , x_{n})\mapsto x_{1}v_{1}+\dots x_{n}v_{n} \end{equation}

is a homeomorphism.

Okay, so my question is: If this is proved and any norm in $V$ is indeed equivalent to the maximum norm in $V$, then why is the isomorphism above a homeomorphism?

I mean, he doesn't even mention which topology we have in $K^{n}$. Is there a canonical one that I should consider? Besides that, he says in the proof: "In fact, $||\: ||$ is transformed into the maximum norm on $K^{n}$", and that really got me lost on this.

Thanks in advance for any explanations.