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This was a many years lasting problem to me, but now that I began to think of it anew, I found the solution: Since $E$ has the Heine−Borel property, taking $V$ to be $\mathcal T\$−closed, we get $S\cap V$ also such, and hence $\mathcal T\$−compact. Having a compact zero neighbourhood, so $S$ must be finite−dimensional. No such complicated things as nuclearity is are needed. The same argument also gives the same result for the space $E=\mathbb R^{{\ \mathbb N}_0}$ .
This was a many years lasting problem to me, but now that I began to think of it anew, I found the solution: Since $E$ has the Heine−Borel property, taking $V$ to be $\mathcal T\$−closed, we get $S\cap V$ also such, and hence $\mathcal T\$−compact. Having a compact zero neighbourhood, so $S$ must be finite−dimensional. No such complicated things as nuclearity is needed.