Let $(d_1,d_2, ...)$ be the sequence obtained by taking the first digit of the first number, the second digit of the first number, etc. Define
$$s: \{0, 1, ... , 9\} \rightarrow \{0, 1, ... , 9\}$$
by $s(x) = x+ 1$ if $x \neq 9$, and $s(9) = 0$. Then your new sequence is
$$(s(d_1), s(d_2), ... )$$
which cannot be on the original countable list.
To me, it's obvious that you can "construct" the above sequence and work with it. If you're not convinced, we can do things more formally as follows. Let $P = \{0, 1, ... , 9\}$. We have the function $s: P \rightarrow P$ I mentioned. Consider the cartesian product
$$\mathbf P = \prod\limits_{i=1}^{\infty} P$$
together with the projections $\pi_i: \mathbf P \rightarrow P, i = 1, 2, ...$. Then every decimal sequence $\mathbf e = (e_1, e_2, ...)$ is an element of $\mathbf P$, and $\pi_i$ is just the map which sends such a sequence to $e_i$. The universal property of the cartesian product guarantees the existence of a unique function
$$\mathbf s: \mathbf P \rightarrow \mathbf P$$
with the property that $\pi_i \circ \mathbf s = s \circ \pi_i$ for every $i$. Then
$$(s(d_1), s(d_2), ... )$$
is nothing more than the function $\mathbf s$ applied to $(d_1, d_2, ...)$.