Given two equivalent norms, convergence in one implies convergences in other. Suppose we have two equivalent norms, $\|\cdot \| \sim \left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert$.
We assume that $x_n \rightarrow x $ in $(X,\|\cdot \|)$. Show that $x_n \rightarrow x $ in $(X,\left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert)$
We observe since the norms are equivalent, there exits constants $\alpha, \beta$, such that $\alpha \|\cdot \| \leq \left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert \leq \beta \|\cdot \|$, with $0 \lt \alpha \leq \beta$
Since, $x_n \rightarrow x$ in $(X,\|\cdot \|$) we know that for all $\epsilon>0, \exists K(\epsilon)$ such that for n>$K(\epsilon),$ we have $\|x_n -x\| \leq \epsilon$
Since, we know $\beta \gt 0$ , then $ \beta \|x_n -x\| \leq \beta \epsilon$,
Thus we see that $\left\vert\!\left\vert\!\left\vert x_n-x\right\vert\!\right\vert\!\right\vert\leq \beta \epsilon$.
This is how far I've gotten, our goal to is to find a a $J(\epsilon)$ such that for all $\epsilon>0, \exists J(\epsilon)$ such that for n>$J(\epsilon),$ we have $\left\vert\!\left\vert\!\left\vert x_n-x \right\vert\!\right\vert\!\right\vert \leq \epsilon$
My candidate for $J(\epsilon$) is $J(\epsilon) = \frac{K(\epsilon)}{\beta}$ from my above argument, but I'm not exactly sure how to validate this. If someone could help me on the proof, it would be very appreciated. I believe I'm pretty close, just missing something small. 

I think you mean $J(\epsilon) = K(\epsilon/\beta)$.
For all $n \ge J(\epsilon)=K(\epsilon/\beta)$, we have $\|x_n-x\| \le \epsilon/\beta$, so
$$||| x_n-x||| \le \beta \|x_n-x\| \le \beta \epsilon/\beta=\epsilon.$$
