Show that $\mathbb R≅\mathbb R \times\mathbb R$ The strategy is to create two injections then apply the Cantor-Berstein Theorem to show their equivalence. Let $\mathbb R$ denote the set of real numbers.
The first injection is $g: \mathbb R\to\mathbb R\times\mathbb R$ given by $h(x)=(x,x),$ where you prove injectivity.
Then the second injection is $f:(0,1)\times(0,1)\to(0,1)$ be defined as $(x,y)$ is an element of $(0,1)\times(0,1),$ we write $x=0.x_1x_2x_3\ldots$ and $y=y_1y_2\ldots$ and have $f(x,y)=0.x_1y_1x_2y_2\ldots,$ how do you prove this is injective?
Then relate $f$ to $R≅(0,1)$ to get the injective function for $\mathbb R\times\mathbb R\to\mathbb R$.
Then you can conclude with the CSB theorem.
 A: The map $f$ is injective. 
Let $x$ have decimal expansion $.x_1x_2x_3\dots$
Let $x^{`}$ have decimal expansion $.x_1^{`}x_2^{`}x_3^{`}\dots$
Let $y$ have decimal eypansion $.y_1y_2y_3\dots$
Let $y^{`}$ have decimal eypansion $.y_1^{`}y_2^{`}y_3^{`}\dots$
For the purpose of standardization, assume that none of these expansions trail off with $999\dots's$ (in the open interval $(0,1)$, this is no easier said than done), and then of course $f$ is well-defined and the form of $f(x,y)$ does not trail off with $999\dots's$ either.
If $x \ne x^{`}$ then $x_k \ne x_k^{`}$ for some $k$.
So $f(x,y) \ne f(x^{`},y^{`})$.
Exercise: Complete this sketchy argument to show that $f$ is injective.
Note that the function $f:(0,1)\times(0,1)\to(0,1)$ i̶s̶ ̶a̶l̶s̶o̶ ̶s̶u̶r̶j̶e̶c̶t̶i̶v̶e̶.̶
A: $\mathbb N^\mathbb N$ can be mapped one-to-one to $\mathbb R$ via this construction
$\require{AMScd}$
\begin{CD}
\mathbb N^\mathbb N @>\text{encoding}>> \mathcal B @>\text{binary dev}>> [0,1[ @>\phi>> ]0,1[ @>2x-1>> ]-1,1[ @>\mathrm{argth}>> \mathbb R
\end{CD}

Where $\begin{cases}\phi(0)=\frac12,\phi(\frac12)=\frac14,\phi(\frac14)=\frac18,..,\phi(\frac1{2^n})=\frac1{2^{n+1}}\quad \mathrm{for}\ n\in\mathbb N \\
\phi(x)=x\quad \mathrm{elsewhere}\end{cases}$
Binary sequences with no trailing $1$, we call them $\mathcal B$, can be mapped one-to-one to $[0,1[$ via the binary developpement $x=\mathtt{0,\overline{b_0b_1b_2...}}$ 
And we can encode sequences of naturals $(u_n)_n\in\mathbb N^\mathbb N$ to $\mathcal B$ this way: $\underbrace{11...1}_{u_0 \text{ times } 1}0\underbrace{11...1}_{u_1 \text{ times } 1}0\underbrace{11...1}_{u_2 \text{ times } 1}0\cdots$
[rem: if $u_i=0$ we simply encode by $0$ ]
The details are given in this post, there are other possible encodings from $\mathbb N^\mathbb N$ to $\mathcal B$ (see for instance the one given by Q the platypus).
Encode each $n_1,n_2,n_3,...∈N^N$ by an inﬁnite sequence of 0s and 1s with inﬁnitely many 0s, and give a proof that $N^N$ is equinumerous with $R$.

Now, with this bijection in the pocket it becomes easy to map $\mathbb R^k\mapsto \mathbb R$ because it is essentially interleaving $k$ sequences of integer numbers $(\mathbb N^\mathbb N)^k\mapsto\mathbb N^\mathbb N$ and then encoding the newly built sequence into $\mathcal B$ like previously.
