So the thing is that frames are the same thing as complete Heyting algebras in terms of objects; but they have different names because of the morphisms.
In particular their description is different because it stresses what those morphisms are expected to preserve.
In particular for frames, we say they have arbitrary joins, finite meets which distribute over the joins; and so implicitly we are saying that a frame morphism is a lattice morphism that preserves arbitrary joind and finite meets; whereas for Heyting algebras we are insisting on the $\implies$ operation, which means implicitly that the morphisms should preserve this.
So even though arbitrary meets do exist in a frame (you can prove that a poset has all joins iff it has all meets), not mentioning them means they are not expected to be preserved under the morphisms. In the example of a topology, the one you mentioned, this is because a finite meet is just an intersection, which is preserved under $f^{-1}$ for $f$ a map, whereas an infinite meet is the interior of that intersection, which need not be preserved under $f^{-1}$ even if $f$ is continuous.
$\Lambda$ is just some arbitrary indexing set. We could phrase this without an indexing set at all, as:
Let $\mathcal{A}$ be a set of sublattices of $L$. Then $$\bigcap_{X\in\mathcal{A}}X$$ (the set-theoretic intersection [of the sublattices in that set]) is also closed under meet and join. etc.
"$\{A_\lambda: \lambda\in\Lambda\}$" is the same as the $\mathcal{A}$ here, except that we've explicitly "named" the elements of $\mathcal{A}$ for easy reference, with $\Lambda$ being the set of "names" we're using.
In this specific context I think the use of an indexing set isn't helpful, but in other contexts it can substantially simplify things.
As to the definition of $[H]$, the point is that we're taking $\mathcal{A}$ to be the set of all sublattices of $L$ containing $H$. The intersection $\bigcap_{X\in\mathcal{A}}X$ is then (by the remark above) a sublattice of $L$, and it clearly contains $H$ and is a subset of every sublattice of $L$ containing $H$. So we have:
Let $H\subseteq L$ and let $\mathcal{A}$ be the set of sublattices of $L$ containing $H$. Then $$[H]:=\bigcap_{X\in\mathcal{A}}X$$ (the set-theoretic intersection of those sublattices) is the smallest sublattice of $L$ containing $H$; that is, $[H]$ is a sublattice of $L$, $H\subseteq [H]$, and whenever $J$ is a sublattice of $L$ with $H\subseteq J$ we have $[H]\subseteq J$.
An aside:
It's worth noting that any set can be "converted" into an indexed set, albeit in a rather silly way: use the set itself as the indexing set! That is, suppose I have a set $\mathcal{S}$. Let $\Lambda=\mathcal{S}$ and for $\lambda\in \Lambda$ - that is, for $\lambda\in S$ - let $S_\lambda=\lambda$. Then $\mathcal{S}=\{S_\lambda: \lambda\in \Lambda\}$.
This is admittedly really really silly, but it does tell us that we can always switch from sets to indexed sets without actually doing anything; it's just an introduction of extra language and symbols, which hopefully(!) improves the presentation. Now like I said, in this specific case I think the use of an indexing set actually just makes things messier, but it is often quite convenient.
(That's not to say that there aren't subtleties around indexed sets. For example, the axiom of choice is equivalent to "For every set $\mathcal{S}$ there is some indexing set $\Lambda$ and some 'naming surjection' $f:\Lambda\rightarrow\mathcal{S}$ - basically, $f$ tells us how $\Lambda$ indexes $\mathcal{S}$ - such that $\Lambda$ is well-orderable." But as long as we just want some indexing set, that's not a problem.)
Best Answer
Your topology $\tau$ is indeed a cocomplete lattice. But notice that the notion of "lattice" only includes the order $\leq$ and the finite meets and joins $\land, \lor$.
In your case, the finite meets are indeed intersections: a finite intersection of open sets is an open set, by definition.
The joins (arbitrary ones) are unions: an arbitrary union of open sets is open, by definition.
But, as you notice, an arbitrary intersection of open sets need not be open: it seems as though $\tau$ is not complete. But it is, because of the theorem you mention. So where did we go wrong ?
Well we went wrong when we went from "$\tau$ is not stable under arbitrary intersections" to "$\tau$ doesn't have arbitrary meets".
Let's see why that goes wrong on an easier example. Consider a lattice with $4$ elements :$a,b,c,d$ where $a\leq b$ and $b\leq c, b\leq d$. This is indeed a lattice (check it if you're not convinced !).
Now let's call it $L$ and let $\land_L$ denote the meet in $L$. In particular we have $c\land_L d= b$. But consider the subset $\{a,c,d\}$: it is a partially ordered set, but it's not stable under $\land_L$: indeed $b$ is not in it. Can we conclude that it's not a lattice ? No, indeed in this subset the meet of $c,d$ exists, and it's $a$, but it doesn't coincide with $\land_L$. T
hat can happen on this level, but it can also happen for complete lattices: that is we may have a complete lattice $L$ with a sublattice $L'$ such that both $L,L'$ are complete and $\land_L = \land_{L'}$ (finite meets !) but arbitrary meets in $L'$ don't need to be the same as in $L$.That's exactly what happens here: you're seeing $\tau$ as a sublattice of $\mathcal{P}(X)$ (power set of $X$, with $\subset$). You notice that the two have the same finite meets; and you notice that (denoting by $\bigwedge_X$ arbitrary meets in $\mathcal{P}(X)$) $\tau$ is not stable under $\bigwedge_X$. Does that mean that $\tau$ is not complete ?
Certainly not; just as above. In fact; if $(O_i)_{i\in I}$ is a family of open sets, then $\bigwedge_{i\in I}O_i$ in $\tau$ (not in $\mathcal{P}(X)$ !) is precisely $\mathrm{Int}(\displaystyle\bigcap_{i\in I}O_i)$. Indeed, this is clearly open, it's clearly included in each of the $O_i$'s; furthermore if $O$ is an open set such that $O\subset O_i$ for each $i$, then $O\subset \displaystyle\bigcap_{i\in I}O_i$ by definition of the intersection, and thus, by definition of the interior, $O\subset \mathrm{Int}(\displaystyle\bigcap_{i\in I}O_i)$: thus this is precisely the definition of a meet: it's a lower bound such that every other lower bound is smaller than it.
In fact, there's nothing special about open sets here. Consider the following : let $L$ be a cocomplete and complete (though this second condition is not necessary) lattice, $L'$ a sublattice of $L$ (that is, $L'$ is a subset that is closed under finite meets and finite joins in $L$) such that arbitrary joins in $L'$ exist, and are the same as those in $L$. Then for any subset $S\subset L'$, the meet of $S$ in $L'$ is the join in $L'$ (and thus in $L$) of $\{x\in L'\mid x\leq \displaystyle\bigwedge_L S\}$; and the proof is exactly the same as above ! Remember that the interior of a set is nothing but the union (join) of all open sets included in it.