Real Numbers are Uncountably Infinite/Proof 2 using Ternary Notation/Lemma

Lemma to Real Numbers are Uncountable: Proof 2 using Ternary Numbers
Let $\left\langle{d_n}\right\rangle$ and $\left\langle{e_n}\right\rangle$ be infinite sequences in $\left\{{0, 1}\right\}$ such that:
 * $\exists m \in \N: d_m \ne e_m$

That is, the sequences $\left\langle{d_n}\right\rangle$ and $\left\langle{e_n}\right\rangle$ are different in at least one term.

Then the ternary representations $D = 0.d_1 d_2 \ldots$ and $E = 0.e_1 e_2 \dots$ represent distinct real numbers.

Proof
Let $\left\langle{d_n}\right\rangle \ne \left\langle{e_n}\right\rangle$.

By the Well-Ordering Principle, there is a smallest $n \in \N_{>0}$ such that $d_n \ne e_n$.

, suppose that $d_n = 0$ and $e_n = 1$.

Let $K = 0.d_1 d_2 \ldots d_{n-1} = \sum_{i \mathop = 1}^{n-1} d_i 3^{-i}$.

Let:
 * $\displaystyle D := K + \sum_{i \mathop = n+1}^\infty d_i 3^{-i}$
 * $\displaystyle E := K + 3^{-n} + \sum_{i \mathop = n+1}^\infty e_i 3^{-i} \ge K + 3^{-n}$

But then:

Thus $D < E$, so $D \ne E$.