Bounding the error.
The error between a continued fraction $[a_0;a_1,a_2,\ldots]$ and its truncation to the rational number $[a_0;a_1,a_2,\ldots,a_n]$ is given by
$$
|[a_0;a_1,a_2,a_3,\ldots] - [a_0;a_1,a_2,\ldots,a_n]|=\left|\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots]}\right) - \left(a_0 + \frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\right|=\left|\frac{[a_1;a_2,a_3,\ldots,a_n]-[a_1;a_2,a_3,\ldots]}{[a_1;a_2,a_3,\ldots]\cdot[a_1;a_2,a_3,\ldots,a_n]}\right| \le \frac{\left|[a_1;a_2,a_3,\ldots,a_n]-[a_1;a_2,a_3,\ldots]\right|}{a_1^2},
$$
terminating with $\left|[a_0;a_1,a_2,a_3,\ldots] - [a_0;]\right|\le 1/a_1$; by iterating this recursive bound we conclude that
$$
\left|[a_0;a_1,a_2,a_3,\ldots] - [a_0;a_1,a_2,\ldots,a_n]\right| \le \frac{1}{a_1^2 a_2^2 \cdots a_n^2}\cdot \frac{1}{a_{n+1}}.
$$
Let $D([a_0;a_1,a_2,\ldots,a_n])$ be the denominator of the truncation $[a_0;a_1,a_2,\ldots,a_n]$ (in lowest terms). Then we have a Liouville number if for any $\mu > 0$, the inequality
$$
a_{n+1} \ge \frac{D([a_0;a_1,a_2,\ldots,a_n])^\mu}{a_1^2 a_2^2 \cdots a_n^2}
$$
holds for some $n$. To give a more explicit expression, we need to bound the growth of $D$.
Bounding the denominator.
Let $D(x)$ and $N(x)$ denote the denominator and numerator of a rational number $x$ in lowest terms. Then
$$
D([a_0;a_1,a_2,\ldots, a_n])=D\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\\ =D\left(\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)=N([a_1;a_2,a_3,\ldots,a_n]),
$$
and
$$
N([a_0;a_1,a_2,\ldots, a_n])=N\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\\ =N\left(a_0+\frac{D([a_1;a_2,a_3,\ldots,a_n])}{N([a_1;a_2,a_3,\ldots,a_n])}\right) = a_0 N([a_1;a_2,a_3,\ldots,a_n]) + D([a_1;a_2,a_3,\ldots,a_n]) \\ = a_0 D([a_0;a_1,a_2,\ldots, a_n]) + D([a_1;a_2,a_3,\ldots,a_n]).
$$
So
$$
D([a_0;a_1,a_2,\ldots,a_n]) = a_1 D([a_1;a_2,a_3,\ldots,a_n]) + D([a_2;a_3,a_4,\ldots,a_n]),
$$
and the recursion terminates with $D([a_0;])=1$ and $D([a_0;a_1])=D(a_0+1/a_1)=a_1$. Since we have $D([a_0;a_1,a_2,\ldots,a_n]) \ge a_1 D([a_1;a_2,a_3,\ldots,a_n])$, we can say that $D([a_1;a_2,a_3\ldots,a_n]) \le \frac{1}{a_1}D([a_0;a_1,a_2,\ldots,a_n])$, and so
$$
D([a_0;a_1,a_2,\ldots,a_n]) \le \left(a_1 +\frac{1}{a_2}\right) D([a_1;a_2,a_3,\ldots,a_n]) \le (a_1 + 1)D([a_1;a_2,a_3,\ldots,a_n]).
$$
An explicit bound on the size of the denominator is therefore
$$
D([a_0;a_1,a_2,\ldots,a_n]) \le (a_1+1)(a_2+1)\cdots(a_n+1).
$$
Conclusion.
We conclude the following theorem:
The continued fraction $[a_0;a_1,a_2,\ldots]$ is a Liouville number if, for any $\mu > 0$, there is some index $n$ such that $$a_{n+1} \ge \prod_{i=1}^{n}\frac{(a_i + 1)^{\mu}}{a_i^2}.$$
If $x$ is some algebraic number, then there is come $c>0$ and $n\in\mathbb N$ such that
$$
\left|x-\frac pq\right|\ge{c\over q^n}.\tag1
$$
for all coprime integers $p,q$ such that $x\ne p/q$.
However, it follows from the definition of Liouville number that for all $m\in\mathbb N$ there exists infinitely many pairs of coprime $p,q$ such that
$$
0<\left|x-\frac pq\right|<{1\over q^m}.
$$
If $m$ is large enough such that $q^{-m}<cq^{-n}$. Then, we see that (1) is violated, which indicates that $x$ cannot be algebraic.
Proof of (1)
Since $x$ is algebraic, it must be a root of some polynomial $P(t)$ with integer coefficients:
$$
P(t)=a_nt^n+a_{n-1}t^{n-1}+\dots+a_0.
$$
Because $p$ has finitely many roots, we can find some $k>0$ such that $p(t)\ne0$ whenever $0<|t-x|<k$. Let $M$ be the maximum of $P'(t)$ in $[x-k,x+k]$, so it follows from the mean value theorem that $\exists c\in(x-k,x+k)$ such that
$$
|P(t)|=|t-x|\cdot|p'(c)|\le M|t-x|\tag2
$$
When $t=p/q$ and $0<|x-t|<k$. By the properties of integers, we know
$$
q^n|P(t)|=|a_np^n+a_{n-1}p^{n-1}q+\dots+a_0q^n|\ge1.\tag3
$$
Combining (2) and (3), we get
$$
\left|x-\frac pq\right|\ge{|P(p/q)\over M}\ge{M^{-1}\over q^n}.
$$
Best Answer
To see that $\alpha$ must be irrational, remember that rational numbers always have eventually periodic decimal (or indeed any base) expansions. But $\alpha$ clearly doesn't.
For question $2$, compare the denominators: for any $m$, when $k$ is sufficiently large we have $k!\cdot m<(k+1)!$, giving in turn $${1\over k!\cdot m}>{1\over (k+1)!}.$$ Now first imagine $c=c(\alpha)$, and then note that even if $c\not=c(\alpha)$ we still eventually get a contradiction as $k$ increases.