Observing it is enough to define an injective function: $$f:(0,1)^{\mathbb{N}}\to \mathbb{R}$$
For $\alpha \in (0,1)^{\mathbb{N}}$, we have $\alpha:\mathbb{N} \to (0,1)$.
We define $$f(\alpha)=\sum_{n\geq0}\frac{2\alpha(n)}{3^n}$$
We need to show $f$ is injective by showing $f(\alpha)\neq f(\beta)$ for $\alpha \neq \beta$.
Suppose $\alpha \neq \beta$. Let $k$ be the minimal natural number with $\alpha(k) \neq \beta(k)$. We can assume, without losing generality, $\alpha(k)=0, \beta(k)=1$.
$$f(\alpha)=\sum^{k=1}_{n=0}\frac{2\alpha(n)}{3^n}+\frac{2\alpha(k)}{3^k}+\sum^{\infty}_{n=k+1}\frac{2\alpha(n)}{3^n}$$
$$\leq \sum^{k=1}_{n=0}\frac{2\alpha(n)}{3^n}+\sum^{\infty}_{n=k+1}\frac{2}{3^n}$$
$$=\cdot\cdot\cdot +\frac{2}{3^{k+1}}\cdot\bigg(1+\frac{1}{3}+\frac{1}{3^2}+...\bigg)= \frac{3}{2}$$
$$\cdot \cdot \cdot + \frac{1}{3^k}$$
$$<\sum^{k=1}_{n=0}\frac{2\alpha(n)}{3^n}+\frac{2\beta(k)}{3^n}$$ Since $(\beta(k)=1)$
$$\leq\sum^{k=1}_{n=0}\frac{2\beta(n)}{3^n}+\frac{2\beta(k)}{3^k}+\sum^{\infty}_{n=k+1}\frac{2\beta(n)}{3^n}$$
$f(\alpha) < f(\beta)$, as required.
Is this proof of the uncountability of the set of real numbers okay? Tell me if I went wrong anywhere. Thanks in advance.
Here is a simpler injection into R.
For any binary sequence s that does not end in 1,1,1,1....
map s to $\sum_n s(n)/2^n.$
For any binary sequence s that does end in 1,1,1,1....
map s to 64 + $\sum_n s(n)/2^n.$