Intersection of equivalence relations is an equivalence relation Theorem:

Given a collection $S$ of equivalence relations on a set $X:=\left\{x_{1},...,x_{n}\right\}$, then the intersection of elements in $S$ is also an equivalence relation.

Proof (my try):
The intersection is not empty since all of the equivalence relations should be reflexive , in other words the identity relation $Id_X$  is contained in all of the elements in $S$, hence the least intersection that they can have is $Id_x$ which is itself an equivalence relation, but if it's not the intersection then there exist at least another ordered pair $\left(x_{i},x_{j}\right)$ in $Id_X$, but symmetry implies $\left(x_{j},x_{i}\right)$ $\left(1\le i<j\le n\right)$ is also  contained in $Id_X$, on the hand the other case would be that these ordered pairs by transitivity add another ordered pair, in all of these cases the relation is an equivalence.
Is my proof right?
Have someone any better proof?
 A: I cannot say whether is right or wrong, since I don't understand what you wrote, other than the assertion that the intersection must contain $\operatorname{Id}_X$ and therefore it must be reflexive (which is correct, by the way).
Let $I$ be the intersection. Then $I$ is symmetric because if $x\mathrel Iy$, then $(x,y)$ belongs to every element of $S$. But every element of $S$ is symmetric and therefore $(y,x)$ also belongs to every element of $S$. In other words, $y\mathrel Ix$.
And $I$ is transitive to. If $x\mathrel Iy$ and $y\mathrel Iz$, then both pairs $(x,y)$ and $(y,z)$ belong to each element of $S$. Sense every element of $S$ is transitive, $(x,z)$ also belongs to every element of $S$. So, $x\mathrel Iz$.
A: 
"symmetry implies..." 

Symmetry of what then? 
And also you seem to be saying that $(x_j,x_i)\in\mathsf{Id}_X$ in spite of $i\neq j$. 
I cannot recognize your "proof" as a correct proof.

Let $X$ be a set and let $\mathcal S$ denote the collection of equivalence relations on $X$.
(It is not necessary to demand that $X$ is finite)
Defining $E:=\bigcap\mathcal S$ it must be proved the $E$ is reflexive, symmetric and transitive.


*

*reflexivity.


Every element of $\mathcal S$ is reflexive so that $(x,x)\in E$ for arbitrary $x\in X$.


*

*symmetry.


Let $(a,b)\in E$ or equivalently $(a,b)\in R$ for every $R\in\mathcal S$. Then also $(b,a)\in R$ for every $R\in\mathcal S$ because every $R\in\mathcal S$ is symmetric. This justifies the conclusion that $(b,a)\in E$.


*

*transitivity.


Let $(a,b)\in E$ and $(b,a)\in E$ or equivalently...
Try to do the rest of this yourself, inspired by the handling of symmetry above.
