There are quite a few problems with your argument. First, how do you know that $X$ exists? It does, but you can’t simply assume this: if you want to use the fact, you have to prove it. But the biggest problem comes at the end, when you write:
if I$\in$X' | $\forall X\in X', X\preceq I $ then
the infi({$S_1 , S_2$ })=I
so {$S_1 , S_2$ } has an Infimum
Look again at what you’ve said: if there is some element $I$ of $X'$ with a certain property, then $\{S_1,S_2\}$ has an infimum. But what if no such $I$ exists? You are in fact showing that if $\{S_1,S_2\}$ has an infimum, then $\{S_1,S_2\}$ has an infimum, which is true but doesn’t help you that that $\{S_1,S_2\}$ actually does have an infimum.
The way to prove that $\{S_1,S_2\}$ has an infimum for every $S_1,S_2\in\operatorname{Pt}(A)$ is actually to construct the partition that is the infimum. You want a partition $S$ of $A$ with two properties:
- $S\preceq S_1$ and $S\preceq S_2$, and
- if $T\in\operatorname{Pt}(A)$, and $T\preceq S_1$ and $T\preceq S_2$, then $T\preceq S$.
In order to get (1), you need to ensure that for each $C\in S$ there are $D_1\in S_1$ and $D_2\in S_2$ such that $C\subseteq D_1$ and $C\subseteq D_2$. In other words, there must be $D_1\in S_1$ and $D_2\in S_2$ such that $C\subseteq D_1\cap D_2$. An easy way to ensure this would be to make sure that every $C\in S$ is of the form $D_1\cap D_2$ for some $D_1\in S_1$ and $D_2\in S_2$. This would also make every $C\in S$ as large as it could possibly be, which seems like a good way to try to make (2) hold as well.
So let $S=\Big\{D_1\cap D_2:D_1\in S_1,D_2\in S_2,\text{ and }D_1\cap D_2\ne\varnothing\Big\}\;.$
Now see if you can show that $S$ is a partition of $A$ that satisfies (1) and (2). If you get stuck, I’ll give you more help, but you ought to see how far you can get from here on your own first.