To prove that every chain $\langle P, \leqslant\rangle$ is a lattice, fix some $a, b \in P$ and w.l.o.g assume that $a \leqslant b$. From reflexivity of $\leqslant$ it follows that $a \leqslant a$, hence $a$ is a lower bound of the set $\{a,b\}$. To prove that it is the greatest lower bound note that if some $c \in P$ is another lower bound of $\{a,b\}$ then by the definition of a lower bound we have $c \leqslant a$. It means that $a$ is the greatest lower bound of $\{a,b\}$. Same reasoning shows that $b$ is the least upper bound of $\{a,b\}$.
To prove that every chain $\langle P, \leqslant\rangle$ is distributive, you should just consider all possible relations between three arbitrary elements $a, b, c \in P$ and check that distributive identity holds.
For example, let $a \leqslant b \leqslant c$, hence $a \wedge b = a \wedge c = a$ and $b \vee c = c$, so $$a \wedge (b \vee c) = a \wedge c = a = a \vee a = (a \wedge b) \vee (a \wedge c).$$
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.
Best Answer
Distributivity of addition over multiplication says $$a+bc=(a+b)(a+c)$$ for any $a,b,c$. Setting $b=c=0$ gives $a=a^2$. Setting $b=0$ and $c=1$ gives $a=a(a+1)=a^2+a=a+a$.
(Alternatively, assuming that by "addition distributing over multiplication" you include the case of nullary multiplication where it says $a+1=1$, then you can deduce $a=a+a$ by just dualizing the argument for $a=a^2$. Actually, though, that nullary case does not have to be assumed, since it follows automatically. Indeed, setting $a=1$ and $c=0$ in $a+bc=(a+b)(a+c)$ gives $1=(1+b)1=1+b$ for arbitrary $b$.)