[proofplan]
The equivalence constants compare the two norms pointwise, so they also compare the corresponding distances from any center $x\in V$. We first prove mutual containment of open balls, which gives equality of the two norm topologies. Then we apply the same inequalities to differences $x_n-x$ to compare sequential convergence, and to elements of a subset $S$ to compare boundedness.
[/proofplan]
[step:Use the equivalence constants to compare balls around the same center]
Let $m,M>0$ be constants such that, for every $v\in V$,
\begin{align*}
m\|v\|_a\leq \|v\|_b\leq M\|v\|_a.
\end{align*}
For $x\in V$ and $r>0$, define the $\|\cdot\|_a$-open ball and the $\|\cdot\|_b$-open ball by
\begin{align*}
B_a(x,r)=\{y\in V:\|y-x\|_a<r\}
\end{align*}
and
\begin{align*}
B_b(x,r)=\{y\in V:\|y-x\|_b<r\}.
\end{align*}
Fix $x\in V$ and $r>0$. If $y\in B_a(x,r/M)$, then $\|y-x\|_a<r/M$, and the upper comparison inequality gives
\begin{align*}
\|y-x\|_b\leq M\|y-x\|_a<r.
\end{align*}
Thus $y\in B_b(x,r)$, so
\begin{align*}
B_a(x,r/M)\subset B_b(x,r).
\end{align*}
If $y\in B_b(x,mr)$, then $\|y-x\|_b<mr$, and the lower comparison inequality gives
\begin{align*}
m\|y-x\|_a\leq \|y-x\|_b<mr.
\end{align*}
Since $m>0$, division by $m$ gives $\|y-x\|_a<r$, hence $y\in B_a(x,r)$. Therefore
\begin{align*}
B_b(x,mr)\subset B_a(x,r).
\end{align*}
[guided]
We want to compare the two topologies, and a norm topology is controlled by its open balls. The useful point is that the norm comparison can be applied not only to a vector $v$, but also to the difference vector $y-x$ measuring the distance from $y$ to the center $x$.
Let $m,M>0$ be constants such that, for every $v\in V$,
\begin{align*}
m\|v\|_a\leq \|v\|_b\leq M\|v\|_a.
\end{align*}
For $x\in V$ and $r>0$, define
\begin{align*}
B_a(x,r)=\{y\in V:\|y-x\|_a<r\}
\end{align*}
and
\begin{align*}
B_b(x,r)=\{y\in V:\|y-x\|_b<r\}.
\end{align*}
First fix $x\in V$ and $r>0$. To put a $\|\cdot\|_a$-ball inside the $\|\cdot\|_b$-ball $B_b(x,r)$, we choose the smaller radius $r/M$. If $y\in B_a(x,r/M)$, then
\begin{align*}
\|y-x\|_a<r/M.
\end{align*}
Applying the upper inequality $\|v\|_b\leq M\|v\|_a$ to the vector $v=y-x$, we obtain
\begin{align*}
\|y-x\|_b\leq M\|y-x\|_a<r.
\end{align*}
Hence $y\in B_b(x,r)$, and therefore
\begin{align*}
B_a(x,r/M)\subset B_b(x,r).
\end{align*}
Conversely, to put a $\|\cdot\|_b$-ball inside the $\|\cdot\|_a$-ball $B_a(x,r)$, we choose the radius $mr$. If $y\in B_b(x,mr)$, then
\begin{align*}
\|y-x\|_b<mr.
\end{align*}
Applying the lower inequality $m\|v\|_a\leq \|v\|_b$ to $v=y-x$, we get
\begin{align*}
m\|y-x\|_a\leq \|y-x\|_b<mr.
\end{align*}
Since $m>0$, dividing by $m$ gives
\begin{align*}
\|y-x\|_a<r.
\end{align*}
Thus $y\in B_a(x,r)$, so
\begin{align*}
B_b(x,mr)\subset B_a(x,r).
\end{align*}
[/guided]
[/step]
[step:Deduce equality of the two norm topologies from mutual ball containment]
Let $\tau_a$ denote the topology induced by $\|\cdot\|_a$, and let $\tau_b$ denote the topology induced by $\|\cdot\|_b$.
Let $U\in\tau_b$. For any $x\in U$, by openness in $\tau_b$ there exists $r>0$ such that
\begin{align*}
B_b(x,r)\subset U.
\end{align*}
From the ball containment proved above,
\begin{align*}
B_a(x,r/M)\subset B_b(x,r)\subset U.
\end{align*}
Thus $U$ is open in $\tau_a$, so $\tau_b\subset\tau_a$.
Let $U\in\tau_a$. For any $x\in U$, by openness in $\tau_a$ there exists $r>0$ such that
\begin{align*}
B_a(x,r)\subset U.
\end{align*}
From the ball containment proved above,
\begin{align*}
B_b(x,mr)\subset B_a(x,r)\subset U.
\end{align*}
Thus $U$ is open in $\tau_b$, so $\tau_a\subset\tau_b$. Therefore $\tau_a=\tau_b$.
[/step]
[step:Compare sequential convergence by applying the norm inequalities to differences]
Let $(x_n)$ be a sequence in $V$, and let $x\in V$.
Assume $x_n\to x$ with respect to $\|\cdot\|_a$. This means
\begin{align*}
\lim_{n\to\infty}\|x_n-x\|_a=0.
\end{align*}
For every positive integer $n$, the upper comparison inequality gives
\begin{align*}
0\leq \|x_n-x\|_b\leq M\|x_n-x\|_a.
\end{align*}
Let $\varepsilon>0$. Since $x_n\to x$ with respect to $\|\cdot\|_a$ and $M>0$, there exists a positive integer $N$ such that $\|x_n-x\|_a<\varepsilon/M$ whenever $n\geq N$. Hence $\|x_n-x\|_b<\varepsilon$ whenever $n\geq N$, so $x_n\to x$ with respect to $\|\cdot\|_b$.
Conversely, assume $x_n\to x$ with respect to $\|\cdot\|_b$. Then
\begin{align*}
\lim_{n\to\infty}\|x_n-x\|_b=0.
\end{align*}
For every positive integer $n$, the lower comparison inequality gives
\begin{align*}
0\leq \|x_n-x\|_a\leq m^{-1}\|x_n-x\|_b.
\end{align*}
Let $\varepsilon>0$. Since $x_n\to x$ with respect to $\|\cdot\|_b$ and $m>0$, there exists a positive integer $N$ such that $\|x_n-x\|_b<m\varepsilon$ whenever $n\geq N$. Hence $\|x_n-x\|_a<\varepsilon$ whenever $n\geq N$, so $x_n\to x$ with respect to $\|\cdot\|_a$.
[/step]
[step:Compare boundedness of subsets using the same constants]
Let $S\subset V$.
Assume $S$ is bounded with respect to $\|\cdot\|_a$. Then there exists $R_a\geq 0$ such that, for every $s\in S$,
\begin{align*}
\|s\|_a\leq R_a.
\end{align*}
Using the upper comparison inequality, for every $s\in S$,
\begin{align*}
\|s\|_b\leq M\|s\|_a\leq MR_a.
\end{align*}
Since $MR_a\geq 0$, the set $S$ is bounded with respect to $\|\cdot\|_b$.
Conversely, assume $S$ is bounded with respect to $\|\cdot\|_b$. Then there exists $R_b\geq 0$ such that, for every $s\in S$,
\begin{align*}
\|s\|_b\leq R_b.
\end{align*}
Using the lower comparison inequality, for every $s\in S$,
\begin{align*}
\|s\|_a\leq m^{-1}\|s\|_b\leq m^{-1}R_b.
\end{align*}
Since $m^{-1}R_b\geq 0$, the set $S$ is bounded with respect to $\|\cdot\|_a$.
We have proved equality of the two induced topologies, equivalence of sequential convergence, and equivalence of boundedness. This completes the proof.
[/step]
