Show that $[0,1]$ and $[0,1]\times [0,1]$ have the same cardinality.
I tried solving the problem as follows:
First of all, we try to find an injective mapping from $[0,1]$ to $[0,1]\times [0,1].$ We define $f: [0,1]\to [0,1]\times [0,1],$ such that $f(x)=(x,\frac 12),$ for all $x\in [0,1]$ and $(x,\frac 12)\in [0,1]\times [0,1].$ Clearly, $f$ is an injeective mapping. For if, $x,y\in [0,1]$ such that $f(x)=(x,\frac 12)=f(y)=(y,\frac 12)\implies x=y.$
Next, we try to find an injective mapping from $[0,1]\times [0,1]$ to $[0,1].$ For this, we use the fact the fact that any terminating real numbers say, $0.x_1x_2x_3x_4x_5$ (, where $x_1,x_2,...,x_5\in \{0,1,2,...,9\}$) represents the same real number as $0.x_1x_2x_3x_4x_5'999...,$ where $x_5'=x_5-1,$ provided that $x_5$ is a non-zero digit. For example, $0.234$ represents the same real numbers as $0.233999...$.
Now, we define a function $g:[0,1]\times [0,1]\to [0,1],$ such that if $(x,y)\in [0,1]\times[0,1]$ then we represent $x$ and $y$ in it's decimal form using the above procedure i.e say if $x=\frac 12,$ we write $x$ as $0.4999...$ instead of $0.5,$ and similarly, for $1$ we write it as $0.999...$. However if, $x=0$ then we write the same as, i.e $x=0.00000....$. Then, for every $(x,y)\in [0,1]\times[0,1],$ we define $g((x,y))=0.x_1y_1x_2y_2x_3y_3....$ where $x$ in it's decimal form is represented by $0.x_1x_2x_3x_4...$ and $y$ in it's decimal form is represented by $0.y_1y_2y_3y_4...$. Now, clearly, $g$ is an injective mapping for if say, $g((a,b))=g((c,d)),$ then, $0.a_1b_1a_2b_2...=0.c_1d_1c_2d_2...$ (where, the decimal representation, of $a,b,c,d$ is, $0.a_1a_2a_3...,0.b_1b_2b_3...,0.c_1c_2c_3....,0.d_1d_2d_3...$ respectively) which means $a_1=c_1,b_1=d_1,a_2=c_2,b_2=d_2,$ and so on, and this means $a=c$ and $b=d,$ i.e $(a,b)=(c,d).$
Hence, we have shown that there exists an injective mapping from from $[0,1]$ to $[0,1]\times [0,1]$ and vice-versa. Using, Schröder-Bernstein Theorem, we can conclude, that there exists a bijection between $[0,1]$ and $[0,1]\times [0,1].$ This proves the fact that both $[0,1]$ and $[0,1]\times [0,1],$ have the same cardinality.
I want to know whether my proof is justified or not? I am unsure about the fact about the injective mappings I found are correct or not. If the proof is correct, I want to know if there are any improvements that I can make to improve upon the readability of the proof, or is it good as this. Any suggestions, stating further improvements that can be made to make the proof look "better", would be helpful.
Lastly, if there are any errors in the above proof, please do let me know.