= Solution
The <Banach-Steinhaus theorem>, or <Uniform boundedness principle>, states that for a family $\mathcal T$ of <bounded linear operators> from a <Banach space> $E$ to a normed space $F$, pointwise boundedness $\sup_{T\in\mathcal T}\|Tx\|<\infty$ for every $x\in E$ implies $\sup_T\|T\|<\infty$.
For its proof let $E_n=\{x:\sup_T\|Tx\|\leq n\}$. These are closed and their union is $E$. The <Baire category theorem> gives an $E_n$ with nonempty interior, containing a ball $B(x_0,r)$. For $\|h\|<r$, both $x_0$ and $x_0+h$ belong to $E_n$, so $\|Th\|\leq2n$ for every $T$. Scaling $h$ and taking a limit at the ball boundary gives $\|T\|\leq2n/r$, proving the theorem. The target need not be complete.
For the asserted bounded-set criterion, use the <Banach space> $X^*$, which is complete even if $X$ is not. Each $x\in S$ defines the bounded linear evaluation functional $J_x(f)=f(x)$ on $X^*$. The given hypothesis says this family is pointwise bounded. The <Hahn-Banach theorem> consequence in the question gives $\|J_x\|=\|x\|$, and uniform boundedness yields
$$
\boxed{\sup_{x\in S}\|x\|<\infty.}
$$
This proves that <weak boundedness implies norm boundedness>; the two forms of boundedness coincide for a subset of a normed space.
Finally suppose the two continuous duals for $\|\cdot\|_1$ and $\|\cdot\|_2$ were equal as sets. The unit ball for norm two is bounded under every common continuous functional, using its norm-two continuity. Applying the preceding criterion in norm one bounds that ball in norm one; scaling gives $\|v\|_1\leq C\|v\|_2$. Reversing the roles gives $\|v\|_2\leq C'\|v\|_1$. The norms would be equivalent. Contrapositively, \b[inequivalent norms have different continuous duals], supplying a linear functional continuous for one and discontinuous for the other.
Back to article page
