TITLE: Open sets can't cover a closed interval, so how can every open cover of $[0,1]$ have a finite subcover?
QUESTION [6 upvotes]: The only answer I can come up with to this question is that there are zero open covers for any closed interval, so purely and only in that sense, the interval $[0,1]$ fits the requirement.
But that doesn't seem to me to match the descriptions I see online. Is my answer correct, or have I missed the topological elephant in the room?
POST MORTEM: The elephant, as all of you hinted, was the definition of a subspace $(X_0, \mathscr{O}|X_0)$ of a topological space $(X,\mathscr{O})$. I assumed a naive, incorrect definition in which "open sets" are composed only of intervals contained in $X_0 \subset \mathbb{R}$ that are open on both ends. But as you pointed out, the correct definition, which I somehow skipped over, stipulates the open sets on any topological subspace $(X_0, \mathscr{O}|X_0)$ as $$\mathscr{O}|X_0 := \{U \cap X_0 | U \in \mathscr{O}\}.$$
That means, when $X = \mathbb{R}$ and $X_0 = [0,1]$, that I was missing some really important open sets:

All of $[0,1]$,
$\{[0,n) \ \ | \ 0<n<1\}$,
$\{(m,1] \ \ | \ 0<m<1\}$.

And it's all downhill from there!

REPLY [3 votes]: When an open cover for a subspace $Y \subset X$ is mentioned, its elements are assumed to be open in a subspace topology induced on $Y$ which does not  have the same family of open sets as $X$.
The subspace topology is defined as follows: a set $M$ is open in $Y$ if and only if there is an open (in $X$) $M'$ such that $M = M' \cap Y$. For example, any subspace $Y \subset X$ is open in itself as $Y = Y \cap X$, which gives a trivial open cover for $Y$.
Another example: $\left\{ \left[0; \frac 3 4 \right), \left(\frac 1 4; 1\right]\right\}$ is an open cover for $\left[0; 1\right] \subset \mathbb R$; it is an intersection of  $\left\{ \left(-1; \frac 3 4 \right), \left( \frac 1 4; 2\right)\right\}$ with $\left[0; 1  \right]$.
One can equivalently say that an open cover of $Y \subset X$ is such a family $\left\{ U_\alpha \right\}$ of sets open in $X$ that $Y \subset \bigcup_\alpha U_\alpha$; note the inclusion instead of equality. By intersecting sets in $\left\{ U_\alpha \right\}$ with $Y$ we obtain an open cover of $Y$ in the subspace topology.