I am trying to solve the following problem:
Suppose that $X$ and $Y$ are Riemann surfaces.
Let $\pi:X\rightarrow Y$ be a covering map, and $u:Y\rightarrow \mathbb{R}$ a subharmonic function on $Y$. Show that $u\circ \pi$ is subharmonic on $X$.
Here, we say $\varphi$ is subharmonic on $U$ if any closed disc $D(z,r) \subset U$ of center $z$ and radius $r$ one has $$\varphi(z) \leq \frac{1}{2\pi} \int_0^{2\pi} \varphi(z+ re^{i\theta}) \, d\theta.$$
A covering map is a "local homeomorphism". But without the analytic structure, I have no idea.
And by definition, for every $y\in Y$, we have some neighborhood $U$ of $y$ and the homeomorphism from some component of $f^{-1}(U)\subset X$ to $U$. But here, to show $f\circ u$ is subharmonic, we need to pick the neighborhood of $x\in X$ at first. So the condition is useless for me...
It seems to me that before the notion of a subharmonic function on a Riemann surface really makes sense we need to know that the composition of a subharmonic function and a biholomorphic map is subharmonic. (You might note that the definition of "subharmonic" that you gave really makes sense only in the plane, not an a Riemann surface.)
In any case that's true, and not hard to prove. Hence the answer to your question is yes if $\pi$ is a holoomorphic covering map, since in that case it is locally biholomorphic. It's not clear whether "covering map" is meant to imply holmorphic in this context.
But if $\pi$ is just a topological covering map then what you say you want to prove seems ridiculous - why would it be true? For an actual counterexample let $X=Y=\Bbb C$, $\pi(x,y)=(x,2y)$ and $u(x,y)=x^2-y^2$. (And of course recall that a smooth function in the plane is subharmonic if and only if its Laplacian is non-negative.)