I'm not sure if my proof is completely correct or if the logic is completely sound. Thank you.:
Question: Let $f: X \rightarrow Y$ where $X$ and $Y$ are both normed linear spaces. Suppose that $f$ is continuous. Show that if X and Y are endowed with equivalent norms, then $f$ remains continuous.
Proof: We show the case where both X and Y have equivalent norms and the other case follows. Suppose we have a normed linear space {$X$,$||||_{x_a}$}. We say $||||_{x_b}$ and $||||_{x_a}$ are equivalent meaning: there exists constants $m$, $M$ so that: $m$$||||_{x_a}\leq ||||_{x_b} \leq M||||_{x_a}$ Denote this as Equation 1. Likewise suppose we have a normed linear space {$Y$,$||||_{y_a}$}. We say that $||||_{y_b}$ and $||||_{x_a}$ are equivalent with the same definition above.
Then, we have: $||f(x_1)-f(x_2)||_{y_b} \leq B||f(x_1)-f(x_2)||_{y_a}$ where $B$ is just a constant. Moroever, since we assume that $f$ is continuous on:{$X$,$||||_{x_a}$} to {$Y$,$||||_{y_a}$}, we have by the definition of continuity that: for every $\epsilon > 0$, there exists $\delta$>0, such that: $||f(x_1)-f(x_2)||_{y_a}< \epsilon/B$ , for $||(x_1)-(x_2)||_{x_a}< \delta$. We have from equation 1, that:$||(x_1)-(x_2)||_{x_b} \leq M||(x_1)-(x_2)||_{x_a}$. Thus for $\epsilon>0$, we have $||f(x_1)-f(x_2)||_{y_b} \leq \frac{B\epsilon}{B}$=$\epsilon$ when $||(x_1)-(x_2)||_{x_b}<\delta'$, where $\delta'=M\delta>0$.Thus, proving the result.
I believe there is a mistake in your last line. Rather than using $\Vert x_1 -x_2 \Vert_{x_b} \le M\Vert x_1-x_2 \Vert_{x_a}$, you would want to use the bound $\Vert x_1-x_2 \Vert_{x_a} \le \frac{1}{m}\Vert x_1-x_2\Vert_{x_b}$. This is because we want to bound $\Vert x_1-x_2 \Vert_{x_a}$ from above by some multiple of $\Vert x_1-x_2 \Vert_{x_b}$.
Thus given $x_1 \in X$ and $\varepsilon >0$, we can choose $\delta > 0$ such that $\Vert x_1-x_2\Vert_{x_a} < \delta$ implies that $\Vert f(x_1)-f(x_2)\Vert_{y_a} < \frac{\varepsilon}{B}$. We can then set $\delta'=\delta m $, it then follows that
$$ \Vert x_1-x_2\Vert_{x_b} < \delta' \Rightarrow \Vert x_1-x_2 \Vert_{x_a} \le \frac{1}{m}\Vert x_1-x_2 \Vert_{x_b}<\delta$$
This then implies that $\Vert f(x_1)-f(x_2)\Vert_{y_b} \le B\Vert f(x_1)-f(x_2) \Vert_{y_a}<\varepsilon$. Since the choice of $\varepsilon > 0$ was arbitrary we have that $f$ is continuous at $x_1$. Since $x_1 \in X$ was also arbitrary we have that $f$ is continuous on all of $X$.
One more comment about your proof. It is not true that for all $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x_1,x_2 \in X$, $\Vert x_1-x_2\Vert_{x_a}<\delta$ implies that $\Vert f(x_1)-f(x_2) \Vert_{y_a} < \frac{\varepsilon}{B}$. This would only be true if $f$ was uniformly continuous, since $f$ is only continuous you must first be given $x_1\in X$ and $\varepsilon > 0$ and then you can find a suitable $\delta > 0$.