How to prove the bijection from unit interval to unit-square constructed by Cantor is discontinuous? Can it be proved with mathematical analysis?
 A: If the bijection were (bi)continuous, then the image of the boundary of the square (homeomorphic to a circle) would be a connected closed subset of the interval.  Such a subset is necessarily an interval itself.  But a circle and an interval have distinct fundamental groups: $\pi_1(I)=0$, while $\pi_1(S^1)=\mathbb Z$.  Since the fundamental group is a homeomorphism invariant, the result follows.
A: Here is a more elementary argument.  If there were a continuous bijection $f$ from the interval $[0,1]$ to the square, take the points $f(0)$ and $f(1)$ in the square.  They are necessarily distinct since $f$ is a bijection.  Now take two disjoint paths $\alpha$ and $\beta$ from $f(0)$ to $f(1)$ in the square. By the intermediate value theorem, the image of $\alpha$ in $[0,1]$ must be the entire interval $[0,1]$, and similarly for $\beta$.  Therefore $f$ is not bijective.  The contradiction proves the result.
A: If $f$ is a continuous and bijective function from the unit interval to the unit square it must map compact sets to compact sets. But all closed sets in the unit interval are compact so its maps all closed sets to closed sets. Equivalently it maps open sets to open sets. Thus its inverse $g$ must be continuous as well. Now use any of the other answers ruling out the existence of bicontinuous bijections. 
A: Let $f$ be any continuous bijection from the unit interval to the unit square.
Since, $[0,1]$ is compact $f^{-1}$ will also be a continuous function.
Now consider any point $a$ belonging to $[0,1]$ such that $a$ is an interior point of $[0,1]$ i.e. $a$ belongs to $(0,1)$ then we look at its image $f(a)$. So deleting the point $a$ from $[0,1]$ will give us a disconnected set while its pre-image under $f^{-1}$ i.e. the $unit $ $square$ $-f(a)$ is clearly pathwise connected and hence connected.  (Since, in euclidean spaces pathwise connectedness is equivalent to connectedness).  So, it implies that $f^{-1}$ is a continuous function that maps a connected set to a disconnected set and hence contradiction.
So no bijection between the unit interval and the unit square can be continuous (in particular, cantor bijection!)
A: Note: This does not actually answer the question as stated, since it rules out the possibility of a homeomorphism, not a continuous bijection.  
This proof if a variant of User72694's answer using $\pi_1$. Suppose that we had a homeomorphism from the square interval to the square.  We will first show that their is an interior point of the interval whose image is an interior point of the square. This is easy since any bijection, $\phi$ from the interval to the square restricts to a bijection of the form, $\bar{\phi}:(0,1)\to [0,1]^2-\{\phi(0),\phi(1)\}$. So pick an interior point of $\{\phi(0),\phi(1)\}$ and it's preimage will be an interior point of the interval. Denote the interior point of the interval by $x_0$. Then the bijection $\phi$ indices a bijection, $\hat{\phi}:[0,1]-\{x_0\}\to [0,1]^2-\{\phi(x_0)\}$, which is a homeomorphism if $\phi$ is. But then $[0,1]-\{x_0\}$ is disconnected and  $[0,1]^2-\{\phi(x_0)\}$ is connected, thus their can be no homeomorphism.
Note that this is a variant of the other proof since both proofs rely on some sort of connectivity.  
