Proof (by contradiction) of the emptiness problem I fail to understand the proof of the Emptiness Problem

$E_{TM} = \{\langle M \rangle | M $ is a TM and $L(M) = \emptyset\}$





1) Use the description of $M$ and $w$ to construct $M_1$, which on Input $x$ behaves as follows:


If $x \neq w$, reject
If $x = w$, run $M$ on input $w$ and accept if $M$ does


2) Run $R$ on input $\langle M_1\rangle$ 

3) If $R$ accepts, reject; if $R$ rejects, accept



I do understand the basic idea of a reduction and in particular the reduction of $A_{TM}$ to $Halt_{TM}$, however, 


I do not see how $E_{TM}$ could be used as a subroutine to solve $A_{TM}$. The whole construction of $M_1$ confuses me a lot. To me it looks like $M_1$ is just like a filter that rejects everything except $w$
But why does $M_1$ even have to check if $x$ equals $w$? As soon as $S$ is fed with a particular pair $\langle M,w\rangle$, $x$ will be equal to $w$, no? how can it be anything different than $w$?
$M_1$ rejects everything, including $w$, unless $M$ accepts it (crucial). $M_1$ will then be a machine that always rejects, if and only if $M$ rejects $w$.
The white box treats $M$ and $w$ as constants. They are still something that $S$ received as input, and we know that.
$R$ is a hypothetical machine that decides if its input is a machine that always rejects.
$S$ will receive unknown $\langle M,w \rangle$ as its input. Should $R$ work as intended, then $S$ in fact solves the general problem of deciding if $M$ accepts $w$, which is exactly $A_{TM}$.
It has been proven that $A_{TM}$ is not decidable.

Therefore,

$R$ cannot exist.
Our reference question has more information on the subject. You may find particularly interesting to see how this kind of proof can be generalized (look for Rice's theorem).