Problem on completeness.

$3$ is false due to $[0,1]$ being compact in the usual topology (see Every compact metric space is complete), and $2$ and $4$ are true for the reason you mention (with the metric inherited from $\mathbb R$).

Now to $1$: We could imbue $(0,1)$ with a new metric, not inherited from $\mathbb R$, in which it will be complete, yet which will produce the same topology as usual. Such a metric space will be a subspace of $\mathbb R$ as a topological space but not as a metric space. (Thus the theorem you cited doesn't apply.)

One such example would be $d(x,y)=|f(x)-f(y)|$ where $f(x)=\tan\left(\pi\left(x-\frac{1}{2}\right)\right)$. That would make $(0,1)$ isometric with $\mathbb R$ in the usual sense, and therefore it will be complete.

(Note: $f$ could have been chosen as any homeomorphism of $(0,1)$ to $\mathbb R$, the 'tangent' function is merely a convenience.)

1. True. Take for instance an homeomorphism $T:(0,1)\to \mathbb{R}$ and pull back the metric on $\mathbb{R}$ by setting $d(x,y) = |T(x)-T(y)|$.
2. True. Usual metric.
3. False. As soon as $[0,1]$ is metrized, compactness is equivalent to completeness and total boundedness.
4. True. See point 3.

I do not quite get your flow of thoughts, so I cannot answer where is your misunderstanding. Anyways, completeness has nothing to do with the topology. To clarify a bit why both point 1 and 2 are true, think of the following: if you are just given a space (forget that $(0,1)\subset \mathbb{R}$), then there is nothing that tells you that you are reaching some boundary, because this does not exists in itself! So sequences that converge to zero can be seen also as sequences that diverge.