Position operator acting on wavefunctions. Why is $\left =\hat{X} \left $? Gennaro Auletta in his book makes the following argument to show that multiplicative operator acting on his eigenvectors acts in a multiplicative way on eigenfunctions as well. Here's the argument.
We write the equation for position operator eigenvectors: 
$\hat{X}$ $\left| x \right>$ = $x\left| x \right>$
and since $\hat{X}$ is a hermitian operator we can write:
$\left<{x}\right|\hat{X}$ = $\left<x\right|x$.
Then we multiply the relation above with a state vector $\left|{\psi}\right>$ to get:
$\left<x\right|\hat{X}\left|{\psi}\right> = \left<x\right|x\left|{\psi}\right>$.
Now
$\hat{X}\psi(x) = x\psi(x)$.
My question is: why are we allowed to get the operator $\hat{X}$ out in the last step? 
 A: This is the oldest abuse of notation in the book, namely using the same symbol for operators acting on vectors (kets) as you use for such in a  representation (bras, hence wavefunctions). Physicists are blasé about it, assuming you understand what they mean. 
As per @knzhou 's suggestion, for the purposes of this discussion (only!), I'll use different  symbols for each, $\hat A$ and $\tilde A$, respectively,
$$
\tilde {A} \psi (x) \equiv \langle x| \hat A |\psi \rangle ~.
$$
This is to say 
$$
\tilde A \langle x | \psi \rangle \equiv \int \!\! dy ~~ \langle x| \hat A |y\rangle \langle y |\psi \rangle .
$$
So you may think of $\tilde A$ as a matrix representation of $\hat A$ in x-space (here, but you could equally work in p-space, l,m-space, ...), where you contract over indices (y) of it with those of the "vector in x-space", here the wavefunction.  Assuming one understands what is meant, one conflates the tilde and the caret, and all is fine. Books using this ritual abuse of notation, however, ought to at least throw in an explanatory footnote...
For your particular operator, of course, 
$$
\hat X = \int \!\! dy ~~ |y\rangle  y \langle y |  \\
\tilde X \psi(x)  =  \int \!\! dy ~~ x \delta(x-y) ~ \psi (y) =x \psi(x),
$$
diagonal in this representation. The authors of your text simply stress that $\tilde X$ and $\hat X$ have the same eigenvalues. 
But, of course, the momentum, e.g., is not diagonal, in this representation,
$$
\hat P = \int \!\! dy ~~ |y\rangle  \frac{\hbar}{i} \partial_y \langle y |  \\
\tilde P \psi(x) =   \frac{\hbar}{i} \partial_x~ \psi (x) =-i\hbar \partial_x \psi(x),
$$
and so on.
In real life, one just uses the caret without excessive confusion.
A: Because of the penultimate equation, for any $\psi$, effect of $\hat{X}$ is the same as effect of multiplying the ket by $x$.
However, these operations are heuristic, they are not really justified by such a loose language as "hermitian operator so we can write something involving kets $|x\rangle$"; those kets do not belong to any Hilbert space, one has to go into functional analysis and theory of rigged Hilbert space to justify these things.
