Number of components in a covering space is less than or equal to number of sheets? Context: on p235 of Hatcher's Algebraic Topology, Hatcher proves that if an n-manifold $M$ is connected, it is orientable if and only if its orientable double-cover $\tilde{M}$ has two components.  
The first two lines of the proof are only a statements about covering spaces, but they puzzle me. If we have an 2-sheeted covering space $\tilde{X} \to X$, and $X$ is a connected space, why is it true that $\tilde{X}$ can only have one or two components? I tried to do a proof by contradiction to convince myself, but cannot come up with a suitable one.   
The second problem I'm facing is with the idea conveyed by the second sentence of the proof, which seems to claim that each of the components is mapped homeomorphically to all of $X$.  
Where are these claims coming from? Are they obvious?
 A: In addition to J. Yang's answer that points out how to prove the case where $\tilde{M}$ is 2-sheeted covering space, I thought one might be interested in the more general case in which one has a n-sheeted covering space of a connected and locally connected space X. So here it goes:
Let $p: \tilde{X} \rightarrow X$ be a n-sheeted covering space where $\tilde X$ is locally connected and $X$ is connected. Then the number of path components of $\tilde{X}$ is less than or equal to n.
Proof:
Note that since $\tilde{X}$ is locally connected, its connected components and path components coincide.
Let $C$ be some connected component of $\tilde{X}$. We will prove that the restriction $p:C\rightarrow X$ is surjective.
Let $\tilde{x} \in C$, $y \in X$ and set $x = p(\tilde{x})$.
There exists some path $f:I \rightarrow X$ that connects $x$ and $y$ because X is connected.
By the lifting property, this path lifts uniquely to some path $\tilde{f}:I \rightarrow \tilde X$ that starts at $\tilde x$. Since $\tilde f (I)$ is connected, it must be contained $C$. In particular $\tilde f (1) \in C$ and $p \circ \tilde f (1) = y$. This proves the surjectivity claim.
Now let $C_1, \ldots, C_k$ be the components of $\tilde X$ and $p_j:C_j \rightarrow X$ be the respective restrictions of the covering map. Using surjectivity of each restriction and the fact that $\tilde X$ has n sheets, we get that for every $x \in X$:
$$
k \leq |p_1^{-1}(x)| + \cdots + |p_k^{-1}(x)| = n
$$
$$\tag*{$\square$}$$
And here's a bonus corollary:
Suppose $p: \tilde{X} \rightarrow X$ be a n-sheeted covering space where $\tilde X$ is locally connected and $X$ is connected. If $\tilde X$ has n components, then each component of $\tilde X$ is homeomorphic to $X$ through the restriction of the covering map $p$. In particular, $\tilde X \cong \sqcup_1^n X$.
Proof: Plugging in $k := n$ in the above inequality, you get that $|p_j^{-1}(x)| = 1$ for $x \in X$ and $j = 1, \ldots, n$. This amounts to saying that each restriction $p_j$ is actually bijective. Hence each $p_j$ is a homeomorphism, being a bijective local homeomorphism.
$$\tag*{$\square$}$$
A: Observation: By path lifting property, you will find that any two points connected by a path on the manifold $M$ will be lifted to a path, with a chosen starting point at $\tilde{M}$. So now the whole lifted path stays in exactly one component of $\tilde{M}$.
This means, once you have a preimage $\tilde{p}$ of a point $p \in M$ in a component of $\tilde{M}$, then by connectedness of $M$, since $\forall q$ can be connected by a path from $p$, we will see in this component we can find a complete preimage of $M$ there.
Now you can use the hypothesis that $M$ is two-sheeted. details are left for you, Acton. 
