How to prove $C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$?

I try to make a self-contained answer to the question (more info can be found on this page). The short answer is: yes you are on the right track!

What you are missing:

1) $\|x\|$ is a continuous function with respect to the $\|\cdot\|_\infty$ norm. You can write $x= \sum_m x_m e_m$ with $e_m$ a basis with $\|e_m\|=1$ (similarly, $y= \sum_m y_m e_m$). We are interested in $$\|x-y\| = \left\| \sum_m (x_m - y_m) e_m \right\| \leq \sum_m | (x_m - y_m)| \leq n \| x - y\|_\infty$$ using the triangle inequality. So for every $\epsilon>0$, we can choose a $\delta =\epsilon/n$ such that with $\| x - y\|_\infty < \delta$ it follows that $\|x-y\| < \epsilon$. Therefore, $\|x\|$ is $\infty$-continuous.

2) The unit sphere $S= \{x :\|x\|_\infty = 1\}$ is a compact set thereby $\|x\|$ assumes a minimum $C_1$ and a maximum $C_2$ on it.

If you are interested in an elementary proof, try $C_1 = \min_j \| e_j \|$ and $C_2 = \sum_j \| e_j \|$.