How is $[0,1]$ locally compact if 0 and 1 do not have a neighborhood in $[0,1]$?
The open sets of $[0, 1]$ (in the topology we usually think of*) are the intersections of open sets in $\Bbb R$ with $[0, 1]$. So for instance, because $(-0.3, 0.2)$ is open in $\Bbb R$, we know that $[0, 0.2)$ is open in $[0, 1]$.
* (This is called the "subspace topology", by the way.)
Does that help?
It's important to bear in mind the topology on $[0,1]$, which is the subspace topology inherited from $\mathbb{R}$. Conferring with the definition of the subspace topology, you'll find that $[0,a)$ is an open set in $[0,1]$ for $0<a<1$, though indeed such a set is not open in $\mathbb{R}$.
As an aside, there is a theorem regarding local compactness for which this scenario arises as a special case:
Theorem: Let $X$ be a locally compact Hausdorff space. If $U$ is either an open or a closed subset of $X$, then $U$ is itself locally compact.
Proof: We want to show that any point $x \in U$ has a compact neighborhood $L_x \subset U$. To do this, we begin by noting that the local compactness of $X$ guarantees that $x$ has a compact neighborhood $K \subset X$. This will serve as a starting point.
The intersection of a closed set and a compact set is itself compact, so our strategy going forward will be to find a (finite) collection of closed neighborhoods $\{ C_i \}$ of $x$ such that $\displaystyle K \cap \left( \bigcap_i C_i \right)$ is a proper subset of $U$ (necessarily compact). A good step in the right direction is to take $K \cap \overline{U}$, which prunes away many points in $K \setminus U$.
To get a third closed set $S$ so that $S \cap K \cap \overline{U} \subset U$, we can do the following: note that $\partial U \cap K$ is compact because $\partial U$ is closed (where $\partial U$ denotes the boundary of $U$), and since $X$ is Hausdorff, we can cover $\partial U\cap K$ with open sets that are "far" from $x$. Let $T$ be the union of the open sets contained in the finite subcover this will necessarily admit, and let $S = X \setminus T$ (which is a closed neighborhood of $x$).
$L_x = S \cap K \cap \overline{U}$ is a compact neighborhood of $x$ contained in $U$.
In this scenario, of course, $X = \mathbb{R}$ (a locally compact Hausdorff space), and $U = [0,1]$.