I assume that you are looking at $[0,1]$ and $[0,1]^2=[0,1]\times[0,1]$ as subsets of $\mathbb{R}$ and $\mathbb{R}^2=\mathbb{R}\times\mathbb{R}$ with their standard topologies as metric spaces. I also assume that by "isomorphism", you mean a bi-continuous function between them.
My answer to a question about cardinalities gives a pair of injections between $[0,1)$ and $[0,1)\times[0,1)$ and then cites the Cantor-Bernstein-Schroeder Theorem to get a bijection between $[0,1)$ and $[0,1)\times[0,1)$. Breaking the "finite when possible" rule there and writing $1$ as
$$
\sum_{k=1}^\infty 2^{-k}
$$
we get a bijection between $[0,1]$ and $[0,1]\times[0,1]$. However, this map cannot be bi-continuous, so it is not an isomorphism.
Suppose we had a bicontinuous map $f:[0,1]\leftrightarrow[0,1]\times[0,1]$. Since $f^{-1}$ is continuous, it maps connected sets to connected sets. Let $f\left(\frac{1}{2}\right)=x\in[0,1]\times[0,1]$. Since $f$ is a bijection, $f^{-1}([0,1]\times[0,1]\setminus \{x\})=[0,1]\setminus\{\frac{1}{2}\}$. However, $[0,1]\times[0,1]\setminus \{x\}$ is connected, but $[0,1]\setminus\{\frac{1}{2}\}$ is not.