I was reading the proof of the Rado–Kneser–Choquet Theorem. The statement is there in the image (taken from the book "Harmonic Mapping in the Plane, Duren page-$30$": 
In the proof, he shows that the Poisson integral function $$f(z)=\dfrac{1}{2\pi}\int_0^{2\pi}\dfrac{1-|z|^2}{|e^{it}-z|^2}\phi(e^{it}~dt $$ is locally univalent and then he finishes his proof by using the Argument Principle. Could you please tell me how it can be shown by using the argument principle that the function $f$ is univalent in $\mathbb{D}.$
The theorem you want is on page 9 - the extension of the analytic argument principle to sense preserving harmonic ones; then you just apply it to $f-a$ for any $a$ not in the image of the Jordan curve $f(\mathbb{T})$ since then $f-a$ satisfies the same properties as $f$ (non-vanishing homeomorphism on the boundary etc) and then by the argument principle and discretness of $2\pi N$, $f-a$ has the same number of zeros inside $\mathbb{D}$ when $a$ belongs to the same connected component of $C- f(\mathbb{T})$, so no zeros if $a$ is outside and precisely one zero if $a$ is inside since otherwise $f$ cannot be a homeomorphism at the boundary.
There are two subtle points here, both implied by the sense-preserving property of $f$, first being that zeros are discrete (there are harmonic functions, even harmonic polynomials like $\Re(z)$, with non-discrete zero set) and second is that they all have positive index (otherwise you could have two zeros with index 1 and one with index -1, say since only zeros of analytic functions are apriori guaranteed to have positive index), so if the argument change is $2\pi$ it means there is a unique zero.