About linear bijection between Banach spaces! It is well-known that, by Banach theorem, every continuous, linear and bijective operator between Banach spaces is a isomorphism. There must be a linear bijective and discontinuous operator between Banach spaces! How can we show/construct such a map?
Thanks for all helpings!
 A: Take any infinite-dimensional Banach space and choose its Hamel basis $\left(e_i\right)_{i\in I}$, say, with $\left\|e_i\right\|=1$. Define $f\colon X \to X$ by
$$
f(e_i) = \alpha_i e_i
$$
where ${\alpha_i}$ is any unbounded collection of (nonzero) numbers. Clearly $f$ is a bijection, and $f$ is unbounded, hence discontinuous.
Now, the above relies critically on the Axiom of Choice. If I recall correctly, there was a similar question here, and it's been pointed out that in certain models with negation of Axiom of Choice there are no discontinuous linear maps from Banach spaces to normed linear spaces at all. So there can be no "truly constructive" example.
It's probably a bit off-topic, but such natural examples exist for incomplete normed linear spaces. A well-known one is a derivative operator for $C^\infty\left([0,1]\right)$ with supremum norm. It's not injective, but it is on a quotient by subspace of constant functions.
A: This requires some form of the Axiom of Choice.
Take two Banach spaces  that are not isomorphic, but both have Hamel bases of the same cardinality (which will be the case e.g. if they are both infinite-dimensional and have cardinality $\bf c$) and
define an operator using a one-to-one correspondence between Hamel bases.
A: We can build a bijective linear function which is not continuous even from a Banach space $E$ in itself (clearly $B$ can't be finite dimensional).
Using $AC$ (we need the Hamel basis) it is easy to build a discontinuous linear functional $$\varphi : B \to \mathbb{R}$$ Let $u\in B$ such that $\varphi(u)=1$. Then define $$S : B \to B $$ $$S(x) = x - 2\varphi(x)u$$ such $S$ is linear and bijective, but clearly not continuous. 
linearity 
$S(\alpha x +\beta y ) = \alpha x +\beta y - 2\varphi(\alpha x +\beta y)u = \alpha x -2\alpha\varphi(x)u +\beta y - 2\beta\varphi(y)u = \alpha S(x) + \beta S(y)$
injectivity
$S(x)=0 \Leftrightarrow x- 2\varphi(x)u =0 \Leftrightarrow x=2\varphi(x)u$ But $S(2\varphi(x)u) = -2\varphi(x)u = 0 \Leftrightarrow \varphi(x)=0 $ but again (using the observation just made) $S(x)= x =0$
surjectivity
For every $y\in B$, $ S(y)$ is a preimage of it. In order to prove this result consider this chain of obvious equalities (we are using the definition, the fact that $\varphi(u)=1$ and the linearity of $\varphi$). $$S(S(y))=S(y)-2\varphi(S(y))u=y-2\varphi(y)u-2\varphi(y-2\varphi(y)u)u= y-2\varphi(y)u -2\varphi(y)u+4\varphi(y)u=y$$
$$ \times \times \times \times \times \times $$
I came across this construction when searching for non equivalent complete norm over the same set (so two Banach space over the same set whose norms are not equivalent), in fact $S: B \to B$ induce a new complete norm over $B$. Let's denote with $|| \cdot ||_1$ the original norm, then we have $|| x ||_2 := ||S(x)||_1$. Using linearity, injectivity and surjectivity of $S$ it's not hard to prove the completeness of such norm. And final addendum, the two norm can't be equivalent!
Edit 1 added proof of injectivity surjectivy and linearity
Edit 2 added a factor $2$ in the definition of the operator
Edit 3 added extended calculation as required by OP
