How to show $(\exists x)( \forall y)\varphi\rightarrow( \forall y)(\exists x)\varphi $ is logically valid 
How to show $(\exists x)( \forall y)\varphi\rightarrow( \forall y)(\exists x)\varphi $ is logically valid

Here is my attempt:  
Assume it's not logically valid. Then, there's an interpretation $\mathscr{M}$  for which it's not true. Hence, there's a sequence $\vec a$ in the domain $M$ of $\mathscr{M}$ such that 1) $ \vec a$ satisfies $(\exists x)( \forall y)\varphi$ and 2) $\vec a$ doesn't satisfy $( \forall y)(\exists x)\varphi$ 
1) $\vec a$ satisfies $(\exists x)( \forall y)\varphi$ $\iff$ $\vec a$ doesn't satisfy $(\forall x)( \exists y)\neg\varphi$ $\iff$ $\vec a$ doesn't satisfy $( \exists y)\neg\varphi$ $\iff$ $\vec a$ satisfies $( \forall y)\varphi$ $\iff$ $\vec a$ satisfies $\varphi$ 
2) $\vec a$ doesn't satisfy $( \forall y)(\exists x)\varphi$ $\iff$ $\vec a$ doesn't satisfy $(\exists x)\varphi$ $\iff$ $\vec a$ satisfies $(\forall x)\neg\varphi$ $\iff$ $\vec a$ satisfies $\neg\varphi$
Since $\vec a$ satisfies both $\varphi$ and $\neg\varphi$, which is a contradiction, then the formula must be valid.
Is it correct to show it this way?

Edit:
1) If $\vec a$ satisfies $(\exists x)( \forall y)\varphi$, i.e $\neg \forall x(\neg\forall y(\varphi))$, then we have $\vec a$ does not satisfy $\forall x(\neg\forall y(\varphi))$ . Then, there is at least one sequence $\vec a'$ differing from $\vec a$ in at most the ith component not satisfying $\neg\forall y(\varphi)$.  Then, that means  $\vec a'$ satisfies $( \forall y)\varphi$.
2) If $\vec a$ doesn't satisfy $( \forall y)(\exists x)\varphi$, i.e $( \forall y)\neg(\forall x)(\neg\varphi)$, then there is at least one sequence $\vec a''$ differing from $\vec a$ in at most the jth component not satisfying $\neg(\forall x)(\neg\varphi)$. Then, that means   $\vec a''$ satisfies  $(\forall x)\neg\varphi$.
  Now, we have $\vec a'$ satisfies $( \forall y)\varphi$ and $\vec a''$ satisfies  $(\forall x)\neg\varphi$. Then,  there is at least one sequence $\vec a'''$ differing from $\vec a$ in at most the $i$th and $j$th component satisfying $\varphi$ and satisfying $\neg\varphi$, which is a contradiction, then the formula must be valid.  

This is from the book, so I applied what's written in $2$ actually

 A: You're forgetting to use from item 4 the part that says the sequence differs in at most the $i$th component.
For instance, say you have a sequence $a$ satisfying a formula $\psi = \exists x\forall y(\varphi)$. Mendelson uses only the universal quantifier, then, actually, the above rendering of $\psi$ is only syntactic sugar for $\neg \forall x(\neg\forall y(\varphi))$.
For a sequence $a$ to satisfy $\psi$, it means that, by item 2,  $a$ does not satisfy $\forall x(\neg\forall y(\varphi))$.
For $a$ to not satifsy $\forall x(\neg\forall y(\varphi))$, by item 4, it means that there is at least one sequence $a'$ differing from $a$ in at most the $i$th component (in wich $i$ is the index of the variable $x$) not satisfying $\neg\forall y(\varphi)$.

As for your edit, what you can get is actually


*

*A sequence $a'$ differing from $a$ in at most the $i$th position ($i$ being the index of $x$) satisfying $\forall y (\varphi)$.

*A sequence $a''$ differing from $a$ in at most the $j$th position ($j$ being the index of $y$) satisfying $\forall x(\neg\varphi)$.


for $x$ and $y$ can be different variables.
To obtain a contradiction from this you can use item 4 (from Mendelson) to obtain another sequence, say $a'''$, differing from $a$ in at most the $i$th and $j$th position that will work for both 1. and 2. above.
