I like Jochen Wengenroth's approach, and I think there is a point that it is worth to clarify (maybe just a matter of style). If we want to make a norm out of $A$, we need it to be a balanced set, so we'd like to pass to the bounded absolutely convex set $\overline{\operatorname{co}}\left(A\cup(-A)\right)$ or to $A-A$. Any family of linear operators which is point-wise bounded on $A$ is clearly also point-wise bounded on $A-A$. However these sets are in general not closed, so some care is needed, because a bounded absolutely convex not closed set $B$ in general would not produce a Banach disk on its linear span, and in fact in general the statement itself does not hold on such $B$ (see the example in the initial comment).

A cheap solution to make the argument work smoothly is to use the notion of $\sigma$-convexity (see e.g. this MO thread) which also generalize slightly the statement); in particular, it covers both the case of a closed and an open bounded convex set $A$. Recall that for a subset $A$ of a Banach space $X$ the following easy fact hold:

• If $A$ is $\sigma$-convex, it is bounded;
• If $A$ is $\sigma$-convex, $A-A$ is $\sigma$-convex (hence $\sigma$-absolutely convex);
• If $A$ is $\sigma$-absolutely convex, it is a Banach disk, that is, its Minkowski functional is a Banach norm on the linear span of $A$.

As a conclusion, we can follow Jochen Wengenroth's reduction to the standard Banach-Steinhaus theorem. We thus have: Any family of linear operators on a Banach space, which is point-wise bounded on a $\sigma$-convex set $A$, is also uniformly bounded on $A$.

I like Jochen Wengenroth's approach, and I think there is a point that it is worth to clarify. If we want to make a norm out of $A$, we need it to be a balanced set, so we'd like to pass to the bounded absolutely convex set $\overline{\operatorname{co}}\left(A\cup(-A)\right)$ or to $A-A$. Any family of linear operators which is point-wise bounded on $A$ is clearly also point-wise bounded on $A-A$. However these sets are in general not closed, so some care is needed, because a bounded absolutely convex not closed set $B$ in general would not produce a Banach disk on its linear span, and in fact in general the statement itself does not hold on such $B$ (see the example in the initial comment).

A cheap solution to make the argument work smoothly is to use the notion of $\sigma$-convexity (see e.g. this MO thread) which also generalize slightly the statement); in particular, it covers both the case of a closed and an open bounded convex set $A$. Recall that for a subset $A$ of a Banach space $X$ the following easy facts hold:

• If $A$ is $\sigma$-convex, it is bounded;
• If $A$ is $\sigma$-convex, $A-A$ is $\sigma$-convex and symmetric (that is, $\sigma$-absolutely convex);
• If $A$ is $\sigma$-absolutely convex, it is a Banach disk, that is, its Minkowski functional is a Banach norm on the linear span of $A$.

As a conclusion, we can follow Jochen Wengenroth's reduction to the standard Banach-Steinhaus theorem. We thus have: Any family of linear operators on a Banach space, which is point-wise bounded on a $\sigma$-convex set $A$, is also uniformly bounded on $A$.