Let's note $E=C([0,1],\mathbb{R})$ the Banach space of real continuous funtions from the [0,1] interval with the uniform norm.
Is it possible to show a noncontinuous linear form on $E$ exists without using a basis, i.e. without AC?
Let's note $E=C([0,1],\mathbb{R})$ the Banach space of real continuous funtions from the [0,1] interval with the uniform norm. Is it possible to show a noncontinuous linear form on $E$ exists without using a basis, i.e. without AC? 


No. It's impossible. In certain models of $\sf ZF+DC$ there is a property known as "automatic continuity" for Banach spaces, that means that every linear operator to a normed space is continuous. Such models are, for example, Solovay's model where all sets of reals are Lebesgue measurable, and have the Baire property; and Shelah's model where all sets of reals have the Baire property. Note that Solovay's model require the existence of an inaccessible cardinal, a statement which is unprovable from the usual axioms of set theory, whereas Shelah's model does not require assumptions beyond those of $\sf ZF$ (and was in fact used to show that the consistency of the statement "all sets of reals have the Baire property" requires no additional hypothesis). You can find a nice exposition to Solovay's model in Aki Kanamori's "The Higher Infinite", as well Solovay's original paper "A model of settheory in which every set of reals is Lebesgue measurable". Shelah's result is much more difficult (but more rewarding too, as it omits the requirement of an inaccessible cardinal), and can be found in the seventh section of Shelah's celebrated paper "Can you take Solovay's inaccessible away?". The paper is long and requires quite an understanding of forcing and set theory in order to full appreciate its content. To see why either one of these results imply the answer above, see Kechris' "Classical Descriptive Set Theory", there he proves Pettis theorem stating that if a group homomorphism between Polish groups is Baire measurable then it is continuous. The proof goes through in $\sf ZF+DC$, and while it does not immediately imply the automatic continuity for all Banach spaces, the one in the question is indeed separable and so forms a Polish group. However a nice argument suggested to me by Harvey Friedman is that in $\sf ZF+DC$ continuity and sequential continuity are equivalent. If $T$ was a discontinuous linear operator on a nonseparable space, then we could have found a sequence which witnesses that, and the restriction of $T$ to the separable space generated by that sequence has to be discontinuous as well. Other authors have dealt with the question of automatic continuity (for Banach spaces) in such models. Amongst them are Garnir, Wright and Brunner. You can also find some information in Eric Schechter's "Handbook of Analysis and Its Foundations". 

