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First we have the Lp metric defined as $d_p((x_1,...,x_n),(y_1,...,y_n))=(|x_1-y_1|^p+...+|x_n-y_n|^p)^\frac{1}{p}$. Now we want to show for $x\in \mathbb R^n$, for any $\epsilon>0$, there exist a $\delta$, such that $B_{L_p}(x,\delta)\subset B_{L_q}(x,\epsilon)$, where $p>q$.

The problem arises when expanding that distance function. The metric is generalised mean, which holds the opposite way, where $M_p

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    The theorem you are looking to prove is commonly called "all norms are equivalent in finite dimensional spaces". It shouldn't be hard to look it up. (The gist of it is compactness and continuity: a closed ball in one norm is compact, so another norm, being continuous, takes on its maximum and minimum on that ball.) – 2017-02-15
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    Minor error on my part: when I said "ball" I should have said "sphere". – 2017-02-15

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First notice that you've already proved that $p>q\implies M_p0,\exists \delta>0,B_{\infty}(x,\delta)\subset B_1(x,\epsilon)$ you're done.

Let $x\in\mathbb{R}^n$ and $\epsilon>0$. Take $\delta=\frac{\epsilon}{n}$ so we have that $$\sup|x_i-y_i| =d_{\infty}(x,y)<\delta\implies d_1(x,y)=\sum|x_i-y_i|< \sum \delta=\epsilon$$