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.
P.S. Here is a somewhat more recent reference which is freely available: George Markowsky, Chain-complete posets and directed sets with applications, Algebra Univ. 6 (1976), 53–68 (pdf).
That result was published by J. Mayer-Kalkschmidt and E. Steiner, Some theorems in set theory and applications in the ideal theory of partially ordered sets, Duke Math. J. 2 (1964), 287–289.
Inasmuch as I don't have access to the paper, which is hidden behind a paywall (the result is stated on the first page, which is freely available), the proof below may not be exactly the same as the one in the paper. I write "directed" for "directed upwards".
Lemma. If $D$ is a directed set of cardinality $\kappa$, an infinite cardinal, then $D$ is the union of a chain of directed subsets, each of cardinality less than $\kappa$.
Proof. We consider three cases.
Case 1. $\kappa=\aleph_0$.
Let $D=\{d_n:n\lt\omega\}$ be an enumeration of $D$. We will define a sequence of finite directed sets $D_n$. Let $D_0=\{d_0\}$. For $n\gt0$, having defined $D_{n-1}$, let $u_n$ be an upper bound for $D_{n-1}\cup\{d_n\}$, and let $D_n=D_{n-1}\cup\{d_n,u_n\}$. Then $\{D_n:n\lt\omega\}$ is a chain of finite directed sets whose union is $D$.
Case 2. $\kappa$ is an uncountable regular cardinal.
Let $D=\{d_\alpha:\alpha\in\kappa\}$. For $\beta\in\kappa$, let $D_\beta=\{d_\alpha:\alpha\lt\beta\}$. Then $B=\{\beta\in\kappa:D_\beta\text{ is directed}\}$ is unbounded in $\kappa$, so that $\{D_\beta:\beta\in B\}$ is a chain of directed sets whose union is $D$, and of course $|D_\beta|\lt\kappa$ for each $\beta$. (To see that $B$ is unbounded, given an ordinal $\beta_o\lt\kappa$, consider the limit of a sequence $\beta_0\lt\beta_1\lt\beta_2\lt\cdots\lt\beta_n\lt\cdots\lt\kappa$ where every finite subset of $D_{\beta_n}$ has an upper bound in $D_{\beta_{n+1}}$.)
Case 3. $\kappa$ is a singular cardinal.
Let $D=\bigcup_{\alpha\in\lambda}E_\alpha$ where $|E_\alpha|\lt\kappa$ and $\lambda=\operatorname{cf}\kappa$. Recursively define directed sets $D_\alpha\subseteq D\ (\alpha\in\lambda)$ so that $|D_\alpha|\lt\kappa$ and $E_\alpha\cup\bigcup_{\xi\lt\alpha}D_\xi\subseteq D_\alpha$. Then $\{D_\alpha:\alpha\in\lambda\}$ is a chain of directed sets whose union is $D$.
Theorem. If a family $K$ of sets is closed under unions of chains, then $K$ is closed under directed unions.
Proof. Assuming the contrary, let $D$ be a directed subfamily of $K$ of minimum cardinality whose union does not belong to $K$. Of course $D$ must be infinite. By the lemma, we can write $D=\bigcup_{\alpha\in\lambda}D_\alpha$, where $\{D_\alpha:\alpha\in\lambda\}$ is a chain of directed families, and $|D_\alpha|\lt|D|$ for each $\alpha$. By the minimality of $|D|$, we have $d_\alpha=\bigcup D_\alpha\in K$ for each $\alpha$. But then, since $\{d_\alpha:\alpha\in\lambda\}$ is a chain in $K$, we have $\bigcup D=\bigcup_{\alpha\in\lambda}d_\alpha\in K$, contradicting our assumption that $\bigcup D\notin K$.
Best Answer
Following the tips of GEdgar i've realized that the answer was pretty simple, sometimes we just need some encouragement ^_^
Algebraic closure implies closure under unions of chains
Let $X$ be a chain with $X_1 \subseteq X_2 \subseteq ...$ (using countable index for simplicity).
$C(\bigcup X)=\bigcup\{C(Y)\ |\ \ (Y \subseteq \bigcup X) \land (|Y| \in \mathbb{N})\}$ (for algebraicity)
For every $Y$ we can find an $X_n$ such that $Y \subseteq X_n$, this prove $C(Y) \subseteq C(X_n)=X_n$. Taking the unions from both sides we get: $C(\bigcup X) \subseteq \bigcup X$
The other verse of inclusion is trivial, hence equality is established and $K$ is closed under unions of chains.
Closure under unions of chains implies algebraic closure
$C(X)=C(\bigcup\{Y\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\})=C(\bigcup\{C(Y)\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\})$
$\{C(Y)\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\}$ is upward directed so its union is closed and the algebraic proprerty is proved.