Composition of subharmonic with holomorphic is subharmonic I need to prove the following claim:

PROBLEM: Asuume $U_1,U_2\subseteq\mathbb C$ are domains in $\mathbb C$. 
  Show that if $f:U_1→U_2$ is holomorphic and $u:U_2\to\mathbb R$ is subharmonic and continuous, then  $u∘f:U_1\to\mathbb R$ is subharmonic.

This problem has already been answered here, but I need a proof that does not use the smooth approximation of subharmonic functions. A solution for the case in which $f$ is invertible can be found here.
I'm trying to generalize the latter solution into every $f\in Hol(U_1)$ ($f$ is not necessarily invertible and $u$ is not necessarily differentiable).
The definition of subharmonic function is:

We say that $u:\Omega\to\mathbb R$ is subharmonic if for every $\bar B(z_0,r)\subseteq\Omega $ we have $$ u(z_0)\leq \frac{1}{2\pi} \int_0^{2\pi} u(z_0+re^{it})dt$$

I have also learned the following theorem:

A continuous function $v:\Omega\to\mathbb R$ is subharmonic if and only if for any harmonic function $u:\Omega'\to\mathbb R$ where $\Omega'\subseteq\Omega$ the difference $v-u$ satisfies the maximum principle in $\Omega'$.

And this claim (I'm not sure it's necessary here):

Let $\Omega=B(z_0,r)$. Assume $v:\Omega\to\mathbb R$ is subharmonic and continuous, and $u:\bar\Omega\to\mathbb R$ is continuous in $\bar\Omega$ and harmonic in $\Omega$. Then if $$\forall a\in\partial\Omega. \limsup_{z\to a} v(z) \leq u(a)$$, then $\forall z\in\Omega. v(z)\leq u(z)$.

Any help would be appreciated!
 A: There is a sketch of the proof in Ransford's book Potential Theory on the Complex Plane, Exercise 2.4.2. I haven't found the proof for the first part:

If $u(z)$ is subharmonic on a neighborhood of the origin, then so is $u(z^k)$ for each $k\geq 1$.

So, let us assume that the above fact it is true. It is not difficult to prove the following statement (Exercise 2.4.2.ii in Ransford's book).

If $f$ is holomorphic on a neighborhood of $w$ and $f-f(w)$ vanishes to order exactly $k$ at $w$, then there exists an injective holomorphic function $g$ on a neighborhood of $w$ such that $f(z) = g^k(z)+f(w)$.

In order to prove this, consider the holomorphic function $h(z) = f(z)-f(w)$. As $h$ has a zero of order $k$, it admits the following representation as a power series:
$$h(z) = h_k(z-w)^k + h_{k+1}(z-w)^{k+1}+\cdots,$$
where $h_k\neq 0$. Thus,
$$h(z) =(z-w)^k\phi(z)$$
where $\phi$ is holomorphic in a neighborhood of $w$ and $\phi(w)\neq 0$. Consequently, we can define the $k$-th root of $\phi$:
$$\psi(z) = \phi^{1/k}(z).$$
Now, the function $h$ has the following representation
$$h(z) = [(z-w)\psi(z)]^k,$$
where $(z-w)\psi(z)$ is holomorphic and injective in a neighborhood of $w$. This fact can be easily checked just differentiating $(z-w)\psi(z)$. Finally, $f(z) = [(z-w)\psi(z)]^k + f(w)$, and taking $g(z) = [(z-w)\psi(z)]^k$ the proposition holds.
Now, using the two propositions above we are able to prove that $u\circ f$ is subharmonic. First, assume that $f(w)=0$, and $w$ is a zero of order $k$ (implicitly I am assuming $0\in U_2$). Then, $f(z) = g^k(z)$ for certain holomorphic and injective map $g$. So, $u\circ f(z) = u(g^k(z))$. But $u\circ g$ is subharmonic in a neighborhood of $0$, and by the first proposition $u(g^k(z))$ is also subharmonic.
In case $f(w)\neq 0$, just consider a small disc around $f(w)$ and take a conformal map $\Phi$ from this disc into the unit disc. Then,
$$u\circ f = (u\circ \Phi^{-1})\circ(\Phi\circ f) = u^*\circ f^*,$$
where $u^*$ is subharmonic and $f^*$ is holomorphic. With this representation we are in the same case as before.
