Derivation of position operator in QM

${\hat X} = x$ because we choose to work in a basis of eigenvectors of ${\hat X}$, i.e. wave-functions.

We have a Hilbert space ${\cal H}$ which is a vector space on which we can choose any basis we wish. We most often choose to work in a basis that diagonalizes the position operator ${\hat X}$. Basis states satisfy $$ {\hat X} | x \rangle = x | x \rangle $$ Once such a basis is chosen, any state $|\Psi\rangle$ in ${\cal H}$ can be expanded in it, i.e. we can write $$ | \Psi \rangle = \int dx | x \rangle \Psi(x) $$ The function $\Psi(x)$ is called the wave-function. We can invert this to find $$ \Psi(x) = \langle x | \Psi \rangle $$

Now, what is the meaning of $\big({\hat X} \Psi \big)(x)$? By definition it is the wave-function of the state ${\hat X} | \Psi \rangle$, i.e. it is $\langle x | {\hat X} | \Psi \rangle$. It is then immediately obvious that $$ \langle x | {\hat X} | \Psi \rangle = x \langle x | \Psi \rangle = x \Psi(x) $$ Thus, we find that $\big({\hat X} \Psi \big)(x) = x \Psi(x)$. For this reason, we write for convenience ${\hat X} = x$, but one must remember that this is only true in the coordinate basis.

Momentum operator in coordinate basis: You might then ask how one can derive the expression for the momentum operator ${\hat P}$ in the coordinate basis. This is done as follows.

By definition $\big( {\hat P} \Psi \big)(x) = \langle x | {\hat P} | \Psi \rangle $. Then, $$ \big( {\hat P} \Psi \big)(x) = \int \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | {\hat P} | \Psi \rangle = \int \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | \Psi \rangle p = \int dx' \frac{dp}{2\pi \hbar} \langle x | p \rangle \langle p | x' \rangle \langle x' |\Psi \rangle p $$ Next, using $\langle x | p \rangle = e^{\frac{i}{\hbar} p x }$, we have $$ \big( {\hat P} \Psi \big)(x) = \int dx' \frac{dp}{2\pi \hbar}e^{\frac{i}{\hbar} p (x-x') } p \Psi(x') $$ We now write $$ e^{\frac{i}{\hbar} p (x-x') } p = - i \hbar \frac{\partial}{\partial x}e^{\frac{i}{\hbar} p (x-x') } $$ Using this, we can explicitly perform the integral over $p$ and we find $$ \big( {\hat P} \Psi \big)(x) = - i \hbar \frac{\partial}{\partial x} \Psi(x) $$ Thus, in coordinate basis, we write $$ {\hat P} = - i \hbar \frac{\partial}{\partial x} $$

PS - Of course, we are free to choose any basis we want. We could for instance work in momentum basis. In this basis, we would have $$ {\hat P} = p~, \qquad {\hat X} = i \hbar \frac{\partial}{\partial p} ~. $$

Note that $\hat{X} = x$ is strictly not correct. The left hand side is the operator acting on the Hilbert states of quantum states. The right-hand side is a real number.

The RHS is the position-representation of the LHS. As the OP recognised correctly this is a tautology to some extent, since it is the defining representation of the operator via the eigenvalue equation

$$\hat{X} |x\rangle = x |x\rangle.$$

So the a priori definition is this one with the notion that the $|x\rangle$-states are localised delta-functions, i.e.

$$\langle x'|x\rangle = \delta^{(3)}\left(x-x'\right).$$

Note that the second equation above follows from the first equation (up to a constant) under the assumption that $\hat{X}$ is hermitian and that its spectrum is continuous.

So far this was only some maths, if you would like to read about this in a more precise manner I recommend Quantum Optics in Phase Space by W. Schleich.

So far this was only maths. The important thing which determines the physics is quantisation, which consists of imposing the relation:

$$\left[\hat{X}, \hat{P} \right] = i\hbar$$

which is also how the first relation in the question (i.e. the momentum operator in position representation) is obtained.