Bounded linear map which is not continuous

For every Hausdorff locally convex space $(X,mathcal T)$ the weak topology $sigma(X,X')$ has the same bounded sets as $mathcal T$, hence the identity $id: (X,sigma(X,X')) to (X,mathcal T)$ is bounded but discontinuous whenever, e.g., $(X,mathcal T)$ has a continuous norm. For example, $(X,mathcal T)$ could be an infinite deimensional normed space.

Locally convex spaces $X$ with the property that every bounded linear map on $X$ is continuous are called bornological. Besides metrizable spaces locally convex inductive limits (aka colimits) of metrizable spaces belong to this class.

Edit: A large class of counterexamples is given in the accepted answer by Jochen. My answer does not actually answer the question, but provides a simple condition under which bounded does imply continuous. After googling bornology, it seems that I have described a specific case of a more general implciation. Namely, if every bornivorous subset of $E$ is a neighborhood of $0in E$, then $E$ is bornological. But the existence of any bounded neighborhood of $0$ implies that every bornivorous subset of $E$ is a neighborhood of $0$. However, perhaps there is still some value in stating this coarser implication without any mention of bornology? I don't know.

Suppose $Usubset E$ is a bounded neighborhood of $0in E$ and $T:Erightarrow F$ is a bounded linear map. Let $Vsubset F$ be a neighborhood of $0in F$. Since $T(U)$ is bounded, there is some $epsilon>0$ such that $T(epsilon U)=epsilon T(U) subset V$. Then $$epsilon Usubset T^{-1}(T(epsilon U))subset T^{-1}(V)$$ and $epsilon U$ is a neighborhood of $0in E$. Thus $T^{-1}(V)$ is a neighborhood of $0in E$. This proves that $T$ is continuous at $0in E$ and thus everywhere, under the assumption that $E$ contains a bounded neighborhood of the origin.

This assumption need not be true (as follows at once from Jochen's answer). To see an example that doesn't mention bornology, consider $mathbb R^mathbb{N}$ with the box topology. If $U$ is a neighborhood of $0$, then there exist $a_1,a_2,ldotsin (0,infty)$ such that $Vsubset U$, where $V=prod_i(-a_i,a_i).$ Now let $W=prod_i(-a_i/i,a_i/i).$ Then there is no $epsilon>0$ such that $epsilon Wsubset V$. Hence $V$ cannot be bounded and thus neither can $U$.