Showing that the product and metric topology on $\mathbb{R}^n$ are equivalent

The product topology is induced by this norm. $$\|x\|_{\rm prod} = \max\{|x_k|, 1\le k \le n\}$$ Let us use $\|\cdot\|$ for the Euclidean norm. Then $$\|x\| = \left(\sum_{k=1}^n x_k^2\right)^{1/2}\le \left(\sum_{k=1}^n \|x\|_{\rm prod}^2\right)^{1/2} = \|x\|_{\rm prod}\sqrt{n}.$$

Now for a reverse inequality.
We have $$|x_k |\le \left(\sum_{k=1}^n x_k^2\right)^{1/2}, \qquad 1\le k \le n,$$ so $$\|x\|_{\rm prod} \le \|x\|.$$

The norms are equivalent.

In order to prove that two topologies are equivalent you need to prove that for each point $x$ and each element open set of the first topology $U_1$ such that $x\in U_1$ there exist some some open set of the second topology $U_2$ such that $x\in U_2\subset U_1$ and vice versa.

Take $x\in\mathbb{R}^n$ and take some open set $U_1$ of metric topology that contains $x$. Since $U_1$ is open we can find a ball $B(x,r)\subset U_1$ that conatins $x$. This ball contains the open box $U_2=(x-\delta,x+\delta)\times\ldots\times(x-\delta,x+\delta)$ for $\delta$ small enough (in fact we can take $\delta=n^{-1/2}r$), which is an open set in the product topology and moreover $x\in U_2\subset B(x,r)\subset U_1$. Hence we proved that metric topology is contained in the product topology.

Again take $x\in\mathbb{R}^n$ and take some open set $U_2$ of the product topology that contains $x$. Since $U_2$ is open we can find a box $(x-\delta,x+\delta)\times\ldots\times(x-\delta,x+\delta)$ which contains $x$. This box contains the ball $U_1:=B(x,0.5\cdot\delta)$ i.e. an element of the base of the metric topology and moreover $x\in U_1\subset(x-\delta,x+\delta)^n\subset U_2$. Thus we proved that product topology is contained in the metric topology.

Since both implications are proved we conclude that this topologies are coincide.