Why
$$\mathbb{E}\left(\mathbb{E}\left(y|x\right)|x\right)=\mathbb{E}\left(y|x\right)?$$
Here is what I managed to do:
$$\mathbb{E}\left[\mathbb{E}\left(y|x\right)|x\right]=\mathbb{E}\left[\int yf\left(y|x\right)dy\,|\,x\right]=\int\int yf\left(y|x\right)dy\,f\left(?|x\right)dy=...=\int yf\left(y|x\right)dy$$
As you can see, I don't know what to put instead of the question mark, then, I don't see how this double integral will simplify to the expected result.
$\mathbb E(y\mid x)$ is a random variable that is measurable wrt the $\sigma$-algebra generated by random variable $x$.
For every random variable $z$ that is measurable wrt the $\sigma$-algebra generated by $x$ we have: $$\mathbb E(z\mid x)=z$$
So also: $$\mathbb E(\mathbb E(y\mid x)\mid x)=\mathbb E(y\mid x)$$
edit (without using the words "algebra" and "measurable").
A "suitable" function $f:\mathbb R\to\mathbb R$ exists such that $\mathbb E(y\mid x)=f(x)$
For every "suitable" function $f:\mathbb R\to\mathbb R$ we have:$$\mathbb E(f(x)\mid x)=f(x)$$
So also: $$\mathbb E(\mathbb E(y\mid x)\mid x)=\mathbb E(f(x)\mid x)=f(x)=\mathbb E(y\mid x)$$
Actually "suitable" is "Borel-measurable" in this context, but that term was contaminated.
edit2
Let $X,Y,Z$ be random variables defined on the same probability space.
If the function $\mathbb R\to\mathbb R$ prescribed by $x\mapsto\mathbb E(Z\mid X=x)$ is denoted by $f$ then $\mathbb E(Z\mid X)$ is a notation for random variable $f(X)$.
Now observe what happens if $Z=g(X)$ for some function $g:\mathbb R\to\mathbb R$.
Then $f(x)=\mathbb E(g(X)\mid X=x)=g(x)$.
So $\mathbb E(g(X)\mid X)$ is actually a notation for $f(X)=g(X)$ or more succinctly:$$\mathbb E(g(X)\mid X)=g(X)\tag1$$
If $Z=\mathbb E(Y\mid X)$ then indeed $Z=g(X)$ for some function $g:\mathbb R\to\mathbb R$.
So application of $(1)$ results in:$$\mathbb E(\mathbb E(Y\mid X) \mid X)=\mathbb E(g(X)\mid X)=g(X)=\mathbb E(Y\mid X)$$