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It seems to be easy, though I cannot spot the trick. I feel it should use Banach-Steinhaus.

Given a normed space $X$, Banach space $Y$ and a sequence of bounded linear maps $T_n\colon X\to Y$ having uniformly bounded norms, suppose that $W\subset X$ is a dense set and $(T_n x)_{n=1}^\infty$ converges for each $x\in W$. Must $(T_n x)_{n=1}^\infty$ converge for each $x\in X$?

gely
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    For $y\notin W$, fix $\varepsilon$ and take $x\in W$ which approximates $y$ to $\varepsilon$. Then try to use a $3\varepsilon$-argument to show that $T_ny$ is a Cauchy sequence. – Giuseppe Negro Jun 19 '13 at 18:48

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Have you solved it? Here's a hint.

Let $y\in X$, $y\notin W$. For every fixed $\varepsilon >0$, you can find a $x\in W$ such that $\lVert x-y\rVert\le \varepsilon$. So by the triangle inequality $$\lVert T_ny-T_my\rVert\le \lVert T_n x-T_mx\rVert+\lVert T_n(x-y)\rVert+\lVert T_m(x-y)\rVert.$$ The sequence $(T_nx)$ is Cauchy. Moreover, we have a uniform bound on the operator norm $\lVert T_n\rVert$, say $\lVert T_n\rVert \le M$ for all $n$. Inserting this information into the inequality above you should get something meaningful.

P.S.: This kind of arguments based on the triangle inequality are sometimes called "$3\varepsilon$-arguments", even if in this case the name $(\varepsilon+\frac{2}{M}\varepsilon)$-argument could be more appropriate!

Giuseppe Negro
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