Equivalence relations with $\cap$, $\cup$ and $\circ$ I have the relations $E_1$ and $E_2$.

A reflexive, symmetric and transitive property shows that two relations are equivalent to each other. I need to prove if this is true for the following

a) $E_1 ∩ E_2$

b) $E_1 ∪ E_2$

c) $E_1 ∘ E_2$

I found out that a) is true, but I am not sure about it.
This is my proof for transitivity:

$E_1 ∩ E_2 ⇒ (a, b, c) ∈ E_1 ∧ (a, b, c) ∈ E_2 ⇒ \\((a, b) ∈ E_1 ∧ (b, c) ∈ E_1 → (a, c) ∈ E_1) ∧
((a, b) ∈ E_2 ∧ (b, c) ∈ E_2 → (a, c) ∈ E_2)$ 

Also I found different arguments on the internet, so I'm pretty unsure about this.

Regarding b) $E_1 ∪ E_2$ a friend found out that this is not transitive and they wanted to prove it with an example.
The example is the following:

$A = \{1,2,3\}$

$E_1 = \{(1,2), (2,1), (2,2)\}$

$E_2= \{(1,3), (3,1), (3,3)\}$

$E_1 ∪ E_2 = \{(1,2), (2,1), (2,2), (1,3), (3,1), (3,3)\}$

This is transitive, isn't it? If not, what is it what I did not understand?

And maybe you can help me with with c), as well.
Here are some hints:

a) If $E_1,E_2$ are equivalence relations then so is $E_1 \cap E_2$. The proof is just by "unpacking" the definitions. (More generally, every class defined by universal statements – those of the form $\forall x_1,\ldots,x_n \phi(x_1,\ldots,x_n)$ – is closed under intersection.)

b) When trying to prove that $E_1 \cup E_2$ is transitive, we get stuck in the following case: if $a E_1 b$ and $b E_2 c$ then we have to show that $a E_1 c$ or $a E_2 c$, though there is no reason to believe that this must be the case. This helps you construct a counterexample.

c) When trying to prove that $E_1 \circ E_2$ is symmetric, we get stuck: if $a E_1 x E_2 b$ then there is no particular reason that $b E_1 y E_2 a$ for some $y$. This helps you construct a counterexample.