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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 ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x\in E$ implies ${\sup }_T\Vert T\Vert <\infty$.

For its proof let $E_n=\{x:{\sup }_T\Vert Tx\Vert \le n\}$. 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 $\Vert h\Vert <r$, both $x_0$ and $x_0+h$ belong to $E_n$, so $\Vert Th\Vert \le 2n$ for every $T$. Scaling $h$ and taking a limit at the ball boundary gives $\Vert T\Vert \le 2n/r$, proving the theorem. The target need not be complete.

For the asserted bounded-set criterion, use the Banach space $X^{\ast }$, which is complete even if $X$ is not. Each $x\in S$ defines the bounded linear evaluation functional $J_x(f)=f(x)$ on $X^{\ast }$. The given hypothesis says this family is pointwise bounded. The Hahn-Banach theorem consequence in the question gives $\Vert J_x\Vert =\Vert x\Vert$, and uniform boundedness yieldsThis 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 $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ 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 $\Vert v{\Vert }_1\le C\Vert v{\Vert }_2$. Reversing the roles gives $\Vert v{\Vert }_2\le C^{\prime }\Vert v{\Vert }_1$. The norms would be equivalent. Contrapositively, inequivalent norms have different continuous duals, supplying a linear functional continuous for one and discontinuous for the other.