I am reading an example that gives a sequence and show how two topologies on the vector space containing the sequence will give different limits of the sequence. They conclude without proof: "the two norms cannot be equivalent" (equivalent in the sense $C_1 \|\cdot\|_1 \leq \|\cdot\|_2 \leq C_2 \|\cdot\|_1$). I don't see why differing limits give in-equivalent topologies.
Note: Open Mapping Theorem has not been introduced yet (which I've read can be helpful here -- but am not sure), so I suspect there's some more elementary/topological argument.
Suppose $C_1 \| \cdot \|_1 \le \| \cdot \|_2 \le C_2 \| \cdot \|_1$, where $C_1, C_2 > 0$. If $x_n \to x$ with respect to $\| \cdot \|_1$, then $$0 \le \|x_n - x\|_2 \le C_2 \|x_n - x\|_1 \to 0$$ in $\Bbb{R}$, so by squeeze theorem, $\|x_n - x\|_2 \to 0$. Note also that $\|x_n - x\|_2 \to 0$ in $\Bbb{R}$ is immediately equivalent to $x_n \to x$ with respect to $\|\cdot\|_2$.
Since $\| \cdot \|_1 \le C_1^{-1} \| \cdot \|_2$, we can see that, if $x_n \to x$ with respect to $\|\cdot\|_2$, then $x_n \to x$ with respect to $\| \cdot \|_1$.
Thus, equivalent norms perfectly preserve convergence (i.e. a convergent sequence in one norm will converge to the same limit in the other). Contrapositively, if two norms cause the same sequence to converge to different limits (or to diverge in one, but not the other), then the norms cannot be equivalent.
EDIT: Here's another way to think about the same proof. Since $\| \cdot \|_2 \le C_2 \| \cdot \|_1$, this means that the identity map $I : X \to X$ on our space $X$ is a bounded linear map from $(X, \|\cdot\|_1)$ to $(X, \|\cdot\|_2)$. In particular, $\|Ix\|_2 = \|x\|_2 \le C_2\|x\|_1$. Hence, $I$ is (sequentially) continuous. So, if $x_n \to x$ under $\| \cdot \|_1$, then $Ix_n \to Ix$ under $\|\cdot\|_2$, by sequential continuity.