$X$ is a $\mathbb{K}$ normed linear space. To show: \begin{equation} \cup_{1\le p < \infty}\{x\in \mathbb{K}^n:||x||_{p}<1\} = \{x\in \mathbb{K}^n:||x||_{\infty}<1\} \end{equation}
I have already shown that the 1st set is contained in the second. And for the 2nd part I think we haveto use $lim_{p\rightarrow \infty}||x||_p=||x||_\infty$, some how.
My approch: Let the second set not be a subset of the first. If we concider a point in the second set, we can find a open ball around it which doesn't intersect with the first set since both the sets are open. Now using the limiting condition we can show that the distance between two norms can be made as close as possible by inceasing $p$. Hence such a open set is not possible. Is this okay? or it lacks rigor?
Let $x\in \cup_{1\le p < \infty}\{x\in \mathbb{K}^n:||x||_{p}<1\}$. Then there is $p\ge 1$ such that $\|x\|_p<1$. Therefore, for each $i\in\{1,2,\cdots,n\}$, $$|x_i|\le\|x\|_p<1$$ which implies $\|x\|_\infty\le \|x\|_p<1$. So $x\in\{x\in \mathbb{K}^n:||x||_{\infty}<1\}$. On the other hand, let $x\in\{x\in \mathbb{K}^n:||x||_{\infty}<1\}$. Then for $i\in\{1,2,\cdots,n\}$, $|x_i|<1$. Without loss of generality, let $$ |x_1|\le |x_2|\le\cdots\le |x_n|, x_n\neq0. $$ Then $$ \|x\|_p\le n^{\frac1p}|x_n|. $$ Noting $\lim_{p\to\infty}n^{\frac1p}=1$, there is $p>1$ such that $n^{\frac1p}<\frac{1}{|x_n|}$. With this $p>1$, $$ \|x\|_p\le n^{\frac1p}|x_n|<1. $$ Thus $x\in \cup_{1\le p < \infty}\{x\in \mathbb{K}^n:||x||_{p}<1\}$. So $$ \cup_{1\le p < \infty}\{x\in \mathbb{K}^n:||x||_{p}<1\}=\{x\in \mathbb{K}^n:||x||_{\infty}<1\}.$$