Prove that $R=\{(M_1,M_2)\in P(N)\times P(N)|M_1\cap M_2=S\}$, for a fixed $S\in P(N)$, with $S\neq N$, $S\neq\varnothing$ is transitive

Question

It is not so clear to me whether this relation is transitive. I have two ideas, one - relation transitive, other - non-transitive.

1. I have so proved that you are transitive:

Let $M_1RM_2 \iff M_1 \cap M_2 = S$, $M_2RM_3 \iff M_2 \cap M_3 = S$, for a fixed $S \in P(N)$, with $S \in N$, $S \neq \varnothing$.

$M_2$ occurs in both equations $\implies$ $M_1$, $M_2$, $M_3$ have the same elements in common, since they are linked to $\cap$.

I.e., $M_1 \cap M_2 = S \iff M_1RM_3$, i.e., transitively.

2. But I can also think with the counterexample (maybe I don’t quite understand what “a fixed $S$” is):

Let $M_1 = \{1, 2\}$, $M_2 = \{2, 3\}$, $M_3 = \{3, 4\}$.

Then $M_1 \cap M_2 = \{2\} = S_1$, $M_2 \cap M_3 = \{3\} = S_2$.

But then $M_1 \cap M_3 = \varnothing = S_3$. Not transitive, since $S_3$ should not be empty.

I cannot understand where I make the mistakes.