If $(X,\|\cdot\|_1)$ is strictly convex, then $(X,\|\cdot\|_2)$ is NOT ALWAYS strictly convex, where $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent. A normed vector space $(X,\|\cdot\|)$ is strictly convex if and only if $x \neq 0$ and $y \neq 0$ and $\|x+y\|=\|x\|+\|y\|$ together imply that $x=cy$ for some constant $c>0$.
We have equivalent norms, which means:
$$\exists a,b : a\|\cdot\|_2 \le \|\cdot\|_1\le b\|\cdot\|_2$$
How can these definitions be used for proof?
 A: You don't; you exhibit a norm $\|\cdot \|_2$ that is equivalent to your strictly convex norm $\|\cdot \|_1$ but isn't strictly convex.  Generally speaking you prove positive statements and you provide counterexamples for negative statements.  Trying to prove a negative can be (almost) impossible (for example, how would you prove that not all integers are even if you're not allowed to point to a non-even integer?)
A standard example for this is the $\ell_2$ and $\ell_1$ norms on ${\mathbb R}^n$  Since we're working in finite ($n$) dimensions we have immediately that the norms are equivalent (though if you've not already seen that you should take a moment to prove it; it's not hard).  Geometrically we see the idea immediately: the unit sphere for $\ell_2$ is what we think of as a sphere and is strictly convex, while the unit sphere for $\ell_1$ is a diamond shape and clearly admits line segments on its surface, meaning it cannot be strictly convex.  (This then leads us to look at the 'points' of the diamond and to the notion of extreme points, and then we can start wondering if the convex hull of those extreme points can regenerate our original set....)
So: in $\ell_1({\mathbb R}^2)$ we can take the points $x_1=(0,1)$ and $x_2=(1,0)$ from the unit sphere and see that $\|x_1 + x_2\|_{\ell_1} =2 = \|x_1\|_{\ell_2} +\|x_2\|_{\ell_2}$ but that $x_1 \not= c\cdot x_2$ for any $c\in {\mathbb R}$.  Hence $({\mathbb R}^2, \|\cdot \|_{\ell_1})$ is not strictly convex.
[If you're interesting in these kinds of renormings then this page provides some more directions.]
