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I am thinking there might be an example between the space of compactly supported smooth functions on the real line (chosen because it is non-metrizable under the standard topology for this space of test functions) and $L^{1/2}[0,1]$ (chosen because it is not locally convex).

Wayne
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Let $E$ be an infinite dimensional Banach space or Fréchet space whose dual has uncountable Hamel basis. Let $F$ be the same space endowed with its weak topology. The identity $\operatorname{id} \colon F \to E$ is bounded (every weakly bounded set is strongly bounded by Mackey's theorem - Banach and Fréchet spaces carry their Mackey topology), but not continuous (the strong topology is strictly finer than the weak topology; for Banach spaces it follows directly because every weak neighbourhood of $0$ contains an infinite dimensional subspace, for Fréchet spaces with big enough dual, you can for every countable family $\mathcal{U} = (U_n)$ of weak $0$-neighbourhoods find a continuous linear form $\lambda$ that is not in the span of the forms used to determine the $U_n$, and $\bigcap \mathcal{U}$ then contains a nontrivial subspace not contained in $\ker \lambda$, whence $\{x\colon \lvert \lambda(x)\rvert < 1\}$ does not contain any $U_n$ [that reasoning applies of course also to infinite dimensional Banach spaces, their dual is an infinite dimensional Banach space, hence has uncountable Hamel basis]).

Daniel Fischer
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    As the above answer shows, it is easier to think of continuity in terms of open sets when considering topological (vector) spaces (at least for questions of this sort) than the usual $\epsilon-\delta$ definition when a norm is available (e.g. in Banach spaces). When considering two topological spaces $X$ and $Y$, the stronger the topology is on $X$, the more continuous maps there are from $X$ to $Y$, and conversely, the stronger the topology on $Y$, the fewer continuous maps there are. – Sargera Jul 12 '13 at 01:05
  • Daniel, in Rudin's Functional Analysis, (1.32) shows that bounded implies continuous if the domain is metrizable, so it seems something is wrong with the example here. – Wayne Jul 15 '13 at 05:36
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    @Wayne But the topology on the domain is the weak topology, and that is not metrizable. The topology on the codomain being metrizable doesn't help (much) there. – Daniel Fischer Jul 15 '13 at 11:11
  • @Daniel Forgive my naivete, but any Banach space or Fréchet space is metrizable, right? – Wayne Jul 15 '13 at 17:43
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    @Wayne Yes, but I took $\operatorname{id} \colon F \to E$, where $E$ is a Fréchet space (or even Banach), and $F$ is the space carrying the weak topology of $E$. If $E$ ($= F$ as vector spaces, but not as topological vector spaces) is infinite dimensional, the weak topology is not metrizable. – Daniel Fischer Jul 15 '13 at 17:48
  • @Daniel I see that F above is not normable, because it is not locally bounded, but I do not see that F is not metrizable. – Wayne Jul 15 '13 at 20:37
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    @Wayne Ah, I see, that's a valid concern. It's a general fact, but of course just saying that would be an argument from authority. For Banach spaces it follows immediately from Rudin's Theorem 1.32, since the fact that the weak topology is not normable directly implies that it's different from the norm topology, hence strictly weaker, hence the identity is not continuous, but if the weak topology were metrizale, 1.32 says it would be continuous. But of course a short direct proof would be nice, however, I have to think about one. Can't promise to find a short and nice one, though. – Daniel Fischer Jul 15 '13 at 20:49
  • @Daniel Thank you for the explanation above. It makes sense and answers my original question perfectly. The more general fact I will work through later. – Wayne Jul 15 '13 at 22:08
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    @Wayne Good that you insisted. Actually, for Fréchet spaces that are not normable, it can happen that the weak and strong topology conicide (if the dual is too small, and has a countable Hamel basis; for example $\omega = \prod_\mathbb{N} \mathbb{K}$ with the product topology). – Daniel Fischer Jul 16 '13 at 09:21