You want to prove that, for any set $A$, if $|\mathbb N\times\mathbb R|\ge|A|$, then $|\mathbb R|\ge|A|$.
No bijections are needed. You just have to show that $|\mathbb R|\ge|\mathbb N\times\mathbb R|$; then you will have $|\mathbb R|\ge|\mathbb N\times\mathbb R|\ge|A|$.
In order to show that $|\mathbb R|\ge|\mathbb N\times\mathbb R|$, you need an injection $f:\mathbb N\times\mathbb R\to\mathbb R$. Here is an easy one:
By the way, it can also be shown that $|\mathbb R\times\mathbb R|\le|\mathbb R|$, but that is more difficult and is not needed here. Of course, using the Cantor-Bernstein theorem, from these inequalities you can get the equalities $|\mathbb R|=|\mathbb N\times\mathbb R|=|\mathbb R\times\mathbb R|$.
Since you asked, here's how you can construct an explicit bijection between $\mathbb N\times\mathbb R$ and $\mathbb R$. You can do that by composing a series of bijections like this:
$$\mathbb N\times\mathbb R\to\mathbb N\times(\mathbb Z\times[0,1))\to(\mathbb N\times\mathbb Z)\times[0,1)\to\mathbb Z\times[0,1)\to\mathbb R.$$
The nontrivial steps are constructing a bijection $\mathbb R\to\mathbb Z\times[0,1)$ and a bijection $\mathbb N\times\mathbb Z\to\mathbb Z$.
A bijection $\mathbb R\to\mathbb Z\times[0,1)$ is easy: $x\mapsto(\lfloor x\rfloor,\ x-\lfloor x\rfloor)$. A bijection $\mathbb N\times\mathbb Z\to\mathbb Z$ can obviously be constructed, but I don't know of an elegant way to define one explicitly, so I leave that part as an exercise.