Open sets can't cover a closed interval, so how can every open cover of $[0,1]$ have a finite subcover? 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!
 A: You may be confusing two different notions of "open cover."
If $X$ is a topological space, an open cover of $X$ is a collection of open sets of $X$ whose union is $X$.
If $E$ is a subspace of $X$, then an open cover of $E$ as a subspace of $X$ is a collection of open sets of $X$ whose union contains $E$.
Note that open sets of $E$ are in general not the same thing as open sets of $X$.  For example, $E$ itself is an open set of $E$, but need not be open in $X$.
So there are two definitions of open cover.  They are related, because if $U_i : i \in I$ is a collection of open sets of $X$ whose union contains $E$, then $E \cap U_i : i \in I$ is a collection of open sets of $E$ whose union is $E$.
A: 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.
A: When applying the definition of compactness, we're viewing $[0,1]$ as a topological space in its own right, rather than a subset of $\mathbb R$.
When the topological space is $[0,1]$, sets such as $[0,\frac12)$ are open in $[0,1]$.
This follows from the definition of the subset topology: The open sets in $[0,1]$ are every open subset of $\mathbb R$ intersected with $[0,1]$, and
$[0,\frac12)= (-1,\frac12)\cap[0,1]$.

Alterantively, and giving the same result, we can view for example, this as an open cover of $[0,1]$:
$$ \{ (-1,1), (0,2) \} $$
This covers $[0,1]$ in the sense that its union is a superset of $[0,1]$.
We don't require that the union of the cover equals the set we want it to cover.
