I think "the" conceptual reason why an uncountable compact subset of $K\subset \mathbb{R}$ has the cardinality of the continuum is that it has a "canonical" map onto a space homeomorphic to $[0,1]$.

The complement of $K$ is a union of open intervals: a left ray, a right ray and at most countably many bounded ones.
The canonical map is the quotient map to the space obtained by gluing the end points of these bounded intervals in the complement
(it is easy to see that the equivalence classes of the equivalence relation thus obtained are at most countable).

This answers the question in the sense that this is a "simple reason why...". I am not saying that this is the simplest proof, as one still has to argue that the quotient space is homeomorphic to $[0,1]$ (a choice of such a homeo is certainly not canonical). There are various ways to do this.

---

**Edit:** below I replace a previous argument by a more conceptual one.

Recall that $K$ is assumed uncountable and that the equivalence relation $\sim$ we defined on it has countable fibers. Denote $X=K/\sim$.
Endow it with the quotient topology and qutient order. 
The following is easy.


Observation: $X$ is a connected, separable, compact linearly ordered space which is not a singleton.

We are left to prove the following.

Proposition: Every space $X$ saisfying the properties above is homeomorphic to $[0,1]$.

Fix a countable dense subset $A\subset X$.
The proof of the proposition consists of the combination of the following facts:

1. $X$ is the isomorphic to order completion of $A$.

2. $A$ is order isomorphic to $\mathbb{Q}$.

3. The order completion of $\mathbb{Q}$ is the two points compactification of $\mathbb{R}$.

Note that 1 and 3 are easy. 2, which is not hard either, is sometimes called "the universal property of the ordered space $\mathbb{Q}$".