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.
So essentially you have the following lattice:
where $a$ and $b$ are compact but $a \wedge b$ is not compact.
Notice that you can define a topological space from this lattice as the one whose open sets are, up to isomorphism, the elements of the lattice.
Make $X = \{ u_0, u_1, u_2, \ldots, a, b \}$ and the open sets are
$$\varnothing, U_0, U_1, U_2, \ldots, U, V, W, X,$$
where $U_i = \{ u_0, \ldots, u_i \}$, $U = \bigcup_i U_i, V = U \cup\{a\}$ and $W = U \cup \{b\}$
Best Answer
For (b), if $f:L\to K$ is a lattice isomorphism between complete lattices, then you want to show that $f(\bigvee A) = \bigvee\{f(a):a\in A\}$, for any $A \subseteq L$.
Now, for $a \in A$, we have that $a \leq \bigvee A$, whence $f(a) \leq f(\bigvee A)$. Therefore, $\bigvee\{f(a):a\in A\} \leq f(\bigvee A)$.
Conversely, since $K$ is complete and $f$ is surjective, there exists $a^* \in L$ such that $f(a^*)=\bigvee\{f(a):a \in A\}$.
Since $f(\bigvee A) \geq f(a^*)$ (which we have seen in the previous paragraph) and $f$ is an order-isomorphism, it follows that $a^* \leq \bigvee A$.
For each $b \in A$, we have $$f(b) \leq \bigvee \{f(a) : a \in A\} = f(a^*),$$ yielding $b \leq a^*$ (because $f$ is an order-isomorphism), and so $\bigvee A \leq a^*$.
We conclude that $a^*=\bigvee A$, and thus, $$f(\bigvee A) = f(a^*) = \bigvee\{ f(a) : a \in A \}.$$