$X$ is a connected space iff every pair of points in $X$ is connected in $X$ The "$\Rightarrow$" statement is trivial, just take $X\subseteq X$.
However I can't prove the "$\Leftarrow$" statement. I tried to prove the contrapositive statement:
Suppose $X$ is disconnected, let $X=U\cup V$ where both $U$ and $V$ are open and non-empty. Let $x,y\in X$ and let $C\subseteq X$ be such that $x,y\in C$. It suffices to show that $C$ is disconnected.
Case 1: if $x\in U$ and $y\in V$, then $x\in C\cap U$, $y\in C\cap V$, and we are done.
Case 2: without loss of generality, suppose $x,y\in U$. Then $x,y\in C\cap U$ which is open in $C$.
But then I can't separate $x$ and $y$.
 A: We should be a little careful with quantifiers. The statement we wish to prove is:

If every pair of points in $X$ is contained in a connected subset $C$ of $X$, then $X$ is connected.

The contrapositive is:

If $X$ is disconnected, then there exists a pair of points in $X$ (call them $x, y$), such that all subsets $C$ containing $x$ and $y$ are disconnected.

My key point is that it is only necessary to exhibit one example of such a pair $x$, $y$.
As you pointed out, $X$ is disconnected, so $X = U \cup V$ for some non-empty, disjoint open subsets $U$ and $V$. So let's pick $x$ to be any randomly-selected point in $U$, and $y$ to be any randomly-selected point in $V$. No matter what $x$ and $y$ we select, any $C$ containing both $x$ and $y$ must be disconnected, for the reason you explained: namely, that $C \cap U$ and $C \cap V$ are non-empty, disjoint open subsets of $C$ that cover $C$. Thus the pair $x$, $y$ serves as a perfectly good example for demonstrating the claim, and no further work is required.
