Let $B^2 :=\{(x,y,x) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$. Show $B^2$ is equinumerous to $\mathbb{R}$.
I think there are a few ways to do this, probably some are easier, but I've committed to defining a bijection between $B^2$ and $\mathbb{R}$. Would this prove that the two are equinumerous and is it necessary to add more detail to my proof:
Define a function $f : B^2 \to \mathbb{R}$ as $f(x^2 + y^2 + z^2 = 1)$ = \begin{cases} 0 & x=1, y=0, z=0 \\ x^ {-1} & x≠0 , y≠0, z=0 \\ x & x≠0 , y=0, z≠0 \\ -1 & x=0 , y=1, z=0 \\ -y^ {-1} & x≠0 , y≠0, z≠0 \\ -y & x=0 , y≠0, z≠0 \\ 1& x=0, y=0, z=1 \\ \end{cases}
Therefore, there exists a bijection $f$ from $B^2$ to $\mathbb{R}$, so $B^2$ is equinumerous to $\mathbb{R}$.
That's not a bijection. It is not injective, e.g. $f(a,0,b)=f(0,-a,b)$ for any nonzero $a^2+b^2=1$. It will be rather hard to construct such bijection explicitly, if even possible.
One way of solving this problem is by constructing an injection (instead of bijection) from something of cardinality $|\mathbb{R}|$ to $B^2$. Because $|B^2|$ is at most $|\mathbb{R}^2|=|\mathbb{R}|$, which is true for any subset of $\mathbb{R}^2$. Such injection can be for example
$$f:[0,1]\to B^2$$ $$f(t)=(t,\sqrt{1-t^2},0)$$
which as a bonus can be easily generalized to arbitrary dimension.