I can prove the statement using the natural deduction, but I keep getting confused about this sequent, so it would be very thankful if someone can help me to understand this concept of predicate calculus.
The sequent is absolutely valid, for arbitrary sets of elements x and y, if x and y are equal or if they have common elements.
however, if x and y take different elements in the universe discourse, i.e, x and y are disjoint, then the sequent may be invalid by soundness of the predicate calculus.
so for example, let the universal discourse be u={1,2,3,4} and define H(x) = "x is 1 or x is 2" for x={1,2} and y={3,4}.
Then, H(1)=H(2)=TRUE but H(3)=H(4)=FALSE, hence (Ǝx)H(x) is true, but (Ǝy)H(y) is false which is a contradiction.
No. Restricting the variables like so is saying: $$(\exists x)\, (x\in\{1,2\}\to H(x)) \dashv\vdash (\exists y)\,(y\in\{3,4\}\to H(y))$$ While this is indeed a fallacious statement for that predicate, it's a completely different one to: $$(\exists x)\, (x\in\{1,2,3,4\}\to H(x)) \dashv\vdash (\exists y)\,(y\in\{1,2,3,4\}\to H(y))$$ Which for a universe of discourse $\{1,2,3,4\}$ is summarised to the given: $$(\exists x)\, (H(x)) \dashv\vdash (\exists y)\, (H(y))$$
This concept you need to grasp is that in this statement the $x$ and $y$ are bound variables within the scope of the existential quantifier. They are what is informally known as "dummy variables". They don't have values assigned outside the scope.
$(\exists x)H(x)$ means that "there is something, let us call it $x$, within the domain of discourse which makes predicate $H(x)$ true." This $x$ is just a placeholder symbol. We can alph-replace it with any token which does not occur free in $H$. Thus, "there is something, let us call it $y$, within the domain of discourse which makes predicate $H(y)$ true," is a logically equivalent statement.