For the Riemann sphere, it is the one point compactification of the plane. So yes, rather by definition it is compact.
Generally, a non-compact topological space can be compactified by suitable addition of "points at infinity". Particularly useful ones (besides the one-point compactification above) include the Stone-Cech compactification, which in some sense adds the "most points", and whose existence for any topological space follows from axiom of choice.
Another compactification that often arises is the Penrose/conformal compactification of space-time in general relativity.
A propos your question on hyperbolic plane. Topologically the hyperbolic plane is the same as the normal plane. By considering the Poincare disk model and "adding" the boundary of the disk, you compactify the hyperbolic plane. As mentioned above, any topological space admits some compactification. What's more useful is to consider special compactifications: for example the one-point compactification of the complex plane into the Riemann sphere is conformal relative to the additional geometrical structure on the two surfaces. Similarly the Penrose compactifications respect the geometrical structure on space-times.
Here is the most down-to-earth explanation I can think of: no evil algebraic geometry, no varieties!
Let me denote by $\mathbb P^1$ the Riemann sphere and write $\mathbb P^1=\mathbb C\cup \lbrace\infty \rbrace$.
Suppose you are given a map $f:\mathbb P^1 \to \mathbb P^1$ and you want to investigate it near $a\in\mathbb P^1$.
First case: $a\neq\infty , f(a) \neq \infty$
Then by restriction you get $f_0: U\to \mathbb C$ for some neighbourhood $U$ of $a$ and you can already handle that.
Second case: $a\neq\infty , f(a) =\infty$
The behavior of $f$ at $a$ is the same as that of $g=1/f$ and we have come back to first case.
Example: $f(z)=1/(z-2)^5$, with $a=2$ . Here $g(z)=(z-2)^5$. We say that $f$ takes value $\infty$ with multiplicity $5$ at $z=2$.
Third case: $a= \infty , f(a) =\infty$
Replace $f$ by $h(z)= \frac{1}{f(\frac {1}{z})}$ and you get a function $h:U\to \mathbb C$ whose behaviour at zero is by definition the behaviour of $f$ at $\infty$.
Example: if $f(z)=(z-2)^5$ , then $h(z)=\frac{z^5}{(1-2z)^5}$ and we say that $f$ has multiplicity $5$ at $\infty$, since $\frac{z^5}{(1-2z)^5}$ has a zero of order $5$ at zero.
Fourth case: $a= \infty , f(a) \neq \infty$
Replace $f(z)$ by $f(1/z)$. I'll skip the details.
Important remark
There are no genuine meromorphic maps $\mathbb P^1 \to \mathbb P^1$: every such meromorphic map is actually holomorphic. The same is true for meromorphic maps between compact Riemann surfaces. This is underappreciated and sometimes misunderstood fact. I actually wrote this answer principally to emphasize this subtle point!
Edit In order to address some comments, let me clarify the above remark.
Given a connected open subset $X \subset \mathbb C$ or even an arbitrary Riemann surface $X$ (compact or not, connected by definition) there is a bijective correspondence between :
1) The set of meromorphic functions $f$ on $X$.
2) The set of holomorphic maps $F: X\to \mathbb P^1$ that are not constantly equal to $\infty$
The correspondence associates to $f$ its extension $F$ obtained by decreeing that at a pole $p\in X$ of $f$ we define $F(p)=\infty$.
I have gone into this discussion because Alphonse wrote in his question "meromorphic functions from (sorry, don't know the notation) Riemann Sphere to itself". My point is that you should only talk about holomorphic maps from
from the Riemann sphere to itself.
Finally note that the bijective correspondence above only holds in dimension 1: there is no way you can consider the meromorphic function $f(z,w)=z/w$ on $\mathbb C^2$ as a holomorphic map $F:\mathbb C^2 \to \mathbb P^1$
Best Answer
The Riemann sphere is not just $\mathbb{C} \cup \{ \infty\}$. It this space endowed with a particular topology. You can think of that topology as arising from adding infinities at the end of some "infinitely large" circle, and then collapsing all those infinities to a point. This perspective is highlighted when considering the stereographic projection of $S^2$ onto $\mathbb{R}^2 $ or $ \mathbb{C}$.
Horizontal circles wrapping around the sphere are mapped to circles in the plane. As we push the circles higher on the Riemann sphere, the corresponding circles in the plane grow. At the north pole, we should have an "infinitely large" circle in the plane, but since the north pole is just a point, that "infinitely large" circle is actually just a point, which we call the point at infinity.
From a technical, point-set topology perspective, this is an instance of the Alexandroff one-point compactification. Another way to view the Riemann sphere is as the complex projective line.