Any two norms on finite dimensional space are equivalent Any two norms on a finite dimensional linear space are equivalent.
Suppose not, and that $||\cdot||$ is a norm such that for any other norm $||\cdot||'$ and any constant $C$, $C||x||'<||x||$ for all $x$. Define $||\cdot||''=\sum |x_i|\cdot||e_i||$ (*). This is a norm and attains at least as large values as $||\cdot||$ for all $x$.
Could this be used as part of a proof? That two norms $||\cdot||_1,||\cdot||_2$ are equivalent means that there are $m,M$ such that $m||x||_1 \leq ||x||_2 \leq M||x||_1$ for all $x$. In the above I only say that there cannot be a norm such that there does NOT exist an $M$ such that $||x||\leq M||x||'$ for any other norm $||\cdot||'$, but I'm really not sure that proves the entire assertion.
If this is utter nonsense, some hints would be appreciated, thank you.
(*The $e_i$ form a base and sloppily assumed the space to be real.)

The negation of the statement that all norms on a finite dimensional space are equivalent is different than what you claim. All you can say is that at least one norm exists for which the inequality doesn't hold, not that it doesn't hold for all norms.
So no, your attempt doesn't work.
