Normal subgroup: a problem in Verification of equivalence

I am familiar with various definitions of normal subgroup. The question I am asking is very trivial one; about a way of proving equivalence.

For a subgroup $N$ of $G$, the following are equivalent: (1) $xN=Nx$ for all $x\in G$. (2) $xNx^{-1}=N$ for all $x\in G$.

My proof: $(1)\Rightarrow (2)$ right multiply by $x^{-1}$.

$(2)\Rightarrow (1)$: Right multiply by $x$.

But, the sources (books/notes) I saw verify the above equivalence by

1. showing LHS is subset of RHS in each statement;

2. proving LHS is in RHS by taking element of LHS and showing that it is in RHS.

But, I wonder, isn't it correct, to prove the equivalence in one or two lines (right multiply by $x$ or by $x^{-1}$ accordingly)?

(In short, can't we avoid "element-belonging" verification for proving equivalence?)

TL;DR The coset notation is really good because it plays nicely with the group operation. However, you need to prove that it plays nicely.

The strings of letters $xN$, $Nx$ and $xNx^{-1}$ are just notation. They represent specific sets, but there is no reason to think that the group operation plays nicely with these sets. You argument assumes that the operation works nicely - for example, that when we multiply every element of $xN$ on the right by $x^{-1}$ we get precisely the set $xNx^{-1}$.

Lets write $xN\cdot x^{-1}$ for the set $\{yx^{-1}\mid y\in xN\}$, so your argument is saying that $xN\cdot x^{-1}=xNx^{-1}$. I think it is pretty clear that $xN\cdot x^{-1}\subseteq xNx^{-1}$, but to prove the other inequality we're going to have to think a bit. (Maybe not much, but a bit.)