Let the starting number be $n$. Consider the case where $n < 10$.
$$ \begin{align}
1 &\to 2 \to 4 \\
2 &\to 4 \\
3 &\to 6 \to 12 \to 24 \to 2 \to 4\\
4 \\
5 &\to 10 \to 1 \to 2 \to 4 \\
6 &\to 12 \to 24 \to 2 \to 4\\
7 &\to 14 \to 1 \to 2 \to 4\\
8 &\to 16 \to 32 \to 64 \to 6 \to 12 \to 24 \to 2 \to 4\\
9 &\to 18 \to 36 \to 72 \to 144 \to 14 \to 1 \to 2 \to 4
\end{align} $$
Now consider general $n > 0$, not ending with the digit $0$ (remove all trailing zeros before beginning).
Let a step from $n$ be defined as the sequence to get from $n$ to a number $m$ for which exactly one trailing digit is removed. For convenience, we write this functionally as $s(n) = m$. The next step is a step from $m$, and we can extend this to a sequence of steps.
Now, let $n = 10k + d$, where $k$ is an integer and $0 \leq d < 10$.
If $d = 0$, we remove the trailing 0 to get $s(n) = \frac{n}{10}$ .
If $d \in \{1,2,3,6,8\}$, first consider $d=1$.
Since $s(n) = \frac{(10k+1)4-4}{10} = 4k$, $s(n)$ is even.
Similarly for the other $d$, in each case $s(n)$ is even.
Since $n$ is doubled at most 3 times, $s(n) \leq \frac{8n}{10}$ .
If $d=4$, then $s(n) = \frac{(10k+4)-4}{10} = k$, i.e. $s(n) < \frac{n}{10}$ .
If $d=5$, then $s(n) = \frac{(10k+5)2}{10} = 2k+1 = \frac{n}{5}$.
If $d=7$, then $s(n) = \frac{(10k+7)2-4}{10} = 2k+1$, i.e. $s(n) < \frac{n}{5}$ .
If $d=9$, we need more steps. Consider $n = 10k+9$ .
We double $n$ four times to get a trailing 4, so
$s(n) = \frac{(10k+9)16-4}{10} = 16k+14$, i.e. $s(n) < 1.6n$. The last digit of $s(n)$ is $16k+14$ mod $10$, for which we only need to consider $0 \leq k \leq 9$. Trying all $k$ from 0 to 9, we find $s(n)$ ends with $4,0,6,2,8,4,0,6,2,8$ respectively. In particular, $s(n)$ never ends with 9.
Apply the above iteratively.
Recall $s(n) < 1.6n$, and $s(n) \mod 10 \in \{0,2,4,6,8\}$ .
If $s(n)$ ends with 0 or 4, then $s(s(n)) \leq \frac{s(n)}{10} < 0.16n$, so $s(s(n)) < n$ .
Otherwise $s(n)$ ends with 2, 6 or 8.
Let $m=s(s(n))$. Then $m$ is even and $m \leq 0.8s(n) < 1.28n$.
If $m$ ends with 0 or 4, then $s(m) \leq \frac{m}{10} < 0.128n$, so $s(m) < n$.
Otherwise $s(m)$ is even and $s(m) \leq 0.8m < 1.024n$, so $s(s(m)) \leq 0.8s(m) < n$.
In every case, there is a sequence of steps taking $n$ to an integer strictly less than $n$, unless we arrive at 4, in which case we've arrived. Call this sequence a jump. Since $n \neq 0$, we never jump to zero.
The jumps reduce $n$ monotonically, so we eventually arrive at a single digit, from which the table above shows that . By inspection of the table above, steps from single digits the sequence always terminates at 4.
QED
