Transferring the usual metric from $\mathbb{R}$ to $(0, 1)$ gives us a complete metric space on $(0, 1)$? I'm watching this video here - https://www.youtube.com/watch?v=zcAvVTFUxS8
The lecturer says that $\mathbb{R}$ and $(0, 1)$ under the usual metric are homeomorphic yet $\mathbb{R}$ is complete and $(0, 1)$ is not. Fair enough.
He then says that, due to the homeomorphism, we can 'pull over' the metric from $\mathbb{R}$ to $(0, 1)$ to form a complete metric space space on $(0,1)$...
That doesn't make sense...its still the same "usual metric" under which $(0,1)$ was not complete. So what am I misunderstanding here?
 A: I suspect there is a confusion about what "pull over" means.
There are two natural ways to get a metric on $(0, 1)$ from the usual metric $d$ on $\mathbb{R}$.
The first is restriction: since $(0, 1)\subseteq\mathbb{R}$ we can take the metric $d_{(0, 1)}$, which is just $d$ itself, limited to points in $(0, 1)$. This is the "usual metric" on $(0, 1)$, and the resulting metric space is not complete.
The second is "pulling over," or maybe better, "pushing forward": you fix an injective function $f: \mathbb{R}\rightarrow (0, 1)$, and then define a new metric $$d_f(f(x), f(y))=d(x, y).$$ So to compute the $d_f$-distance between two points in $(0, 1)$, we pull these points back along $f$ and look at what $d$ says. (Note that we "push forward" the metric - we start with a metric on $\mathbb{R}$, and wind up with a metric on $(0, 1)$ - but to compute its values we "pull back.") If $f$ is nice, then $d_f$ is a metric (EXERCISE: what assumptions do we need to make about $f$?); in particular, $d_f$ is a metric if $f$ is a homeomorphism.
The pushed-forward metric $d_f$ may be (in fact, will be) very different from the restricted metric $d_{(0, 1)}$. In particular, if $f$ is a homeomorphism then $(\mathbb{R}, d)$ and $((0, 1), d_f)$ will be isometric (EXERCISE) so if one is complete, the other will be too.

For an example of how the pushed-forward metric is different from the restricted metric: one nice homeomorphism from $\mathbb{R}$ to $(0, 1)$ is the function $f(x)={\arctan(x)+{\pi\over 2}\over \pi}$. Now, pushing forward along $f$, we have $$d_f({1\over 4}, {3\over 4})=d(-1, 1)=2,$$ while of course $$d_{(0, 1)}({1\over 4}, {3\over 4})={1\over 2}.$$
A: I did not watch the video, but I'm rather sure that no, it's not the 'same usual' metric. If it done correctly it will change the distance from any $x \in (0,1) $to the points $0$ and $1$ from a finite value to $\infty$ 
A: If $f:\mathbb{R} \to (0,1)$ is the homeomorphism, what he means is that we can define $d$ a metric on $(0,1)$ by setting $d(x,y)=|f^{-1}(x)-f^{-1}(y)|$. You can check that this is a metric on $(0,1)$ making it complete. 
A: The first metric $d$ (the usual one) on $X:=(0,1)$ is not complete. The second one $d_1$ is.
Even if those two metrics induce two homeomorphic topological spaces, the two metrics may not be equivalent. For instance, $d$ is a bounded metric whereas $d_1$ is not. This forbid to have isometries, lipschitzian equivalence or even uniform equivalence between those two metric spaces and this is the kind of equivalence you need to deduce the conservation of completude.
To sum up, completude is a metric property whereas homeomorphism is a topological equivalence.
