Is there a continuous bijection between an interval $[0,1]$ and a square: $[0,1] \times [0,1]$?

No, such a bijection from the unit interval $I$ to the unit square $S$ cannot exist. Since $I$ is compact and $S$ is Hausdorff, a continuous bijection would be a homeomorphism (see here). But in $I$ there are only two non-cut-points, whereas in $S$ each point is a non-cut-point.

Hint: Consider what happens to the connected $[0,1]$ if the point $\frac12$ is removed. What happens to $[0,1]\times[0,1]$ when $f(\frac12)$ is removed?

Hint: A continuous bijection from a compact space to a Hausdorff space is bicontinous.

No. The easiest was to see this is to first notice that $[0,1]^2\setminus \{x\}$ is connected for any $x \in [0,1]^2$.

It is easier (for me) to work with $\phi = f^{-1}$. However I must show that $\phi$ is continuous. Suppose $y_n \to y$, then I must show that $\phi(y_n) \to \phi(y)$. Let $x_n = \phi(y_n), x = \phi(y)$. One slightly technical way is to show that every subsequence of $x_n$ contains a further subsequence that converges to $x$. From this we will conclude that $\phi$ is continuous.

Suppose $x_{n_k} \to z$. Since $f$ is continuous, we have $y_{n_k} = f(x_{n_k}) \to f(z) = y$. Hence $z = x$. (So, in fact, the entire sequence, not just a subsequence, converges to $x$.) Hence $\phi$ is continuous.

Now suppose $\phi:[0,1]^2 \to [0,1]$ is a continuous bijection. Let $x = \phi^{-1} (\frac{1}{2} )$, then $\phi([0,1]^2\setminus \{x\})$ is connected, however we see that $\phi([0,1]^2\setminus \{x\}) = [0,\frac{1}{2}) \cup (\frac{1}{2},1]$ which is not connected. Hence a contradiction.