Let $f$ be a function $f:[0,1] \to [0,1] \times [0,1]$ now can we find $f$ with following conditions? Let $f$ is a function $f :[0,1] \to [0,1] ×[0,1] $ now can we find $f$ with following conditions ?: 
1- f be continues and one to one .
2- f be continues and onto .
3- f be continues and one to one and onto .
4- f be onto and one to one but not continues .
I think for case 1 we can consider $f(x)=(x,x)$ and  it is clear f is continues and one to one .
for case 2 there is no f with this condition because $ [0,1] \not \cong [0,1]×[0,1] $ because if we remove point c , $0<c<1 $ from close interval [0,1] then we have disconnect set in domain of f , and but for every point in $[0,1]×[0,1]$ we have no disconnect set.
for case 4 because $[0,1] \cong [0,1]×[0,1] $  and have same cardinals then f exist .
 A: Your answers for $(1), (4)$ is correct. 
For $(3)$, if $f : [0, 1] \to [0, 1]^2$ was a continuous bijection, then as $f$ is an injection from a compact to Hausdorff space, it is a homeomorphism onto it's image. $f$ is surjective, thus $f$ is actually a homeomorphism. But as you noted, this is impossible since removing a point from the interior $[0, 1]$ disconnects $[0, 1]$, whereas removing a point from the interior of $[0, 1]^2$ doesn't make it disconnected. Number of connected components is a homeomorphism invariant, thus there cannot be any such homeomorphism. Contradiction. 
As noted in the comments, the creature in $(2)$ does exist. It's called a space-filling curve. Here's an explicit construction: Let $C \to [0,1]$ be a surjective continuous map (these are easy enough to construct: $C$ can be defined by the usual middle-third construction. But at each step glue the endpoints of the intervals formed. Explicitly, take any point in $C$, expand in ternary expansion. It has an expansion with coefficients either $0$ or $2$. Turn it into a binary expansion by dividing by $2$). Then we have a surjective continuous map $C \times C \to [0, 1]^2$. Note that as $C$ is homeomorphic to $\{0, 1\}^\omega$, $C \cong C \times C$. Thus, we have a surjective continuous map $C \to [0, 1]^2$. Once that is done, use Tietze extension theorem to pull up to a surjective continuous map $[0, 1] \to [0, 1]^2$.
Here's a fun generalization of $(2)$. Every compact metric space is continuous image of the cantor set. 
