It's straightforward to prove that for a topological space, connected $\iff $ having one component.
It then follows by contra-positive that disconnected $\iff$ having more than one component.
In the case of a finite number of components of a space $\{C_i\}_{i = 1, n}$ a direct proof $X$ is disconnected is that $X$ = $C_1 \cup (\cup _{i = 2, n} C_i)$. Then with each component being closed, the finite union $\cup _{i = 2, n} C_i$ is a closed set and $X$ is the union of two non-empty disjoint closed sets $C_1$, and $ \cup _{i = 2, n} C_i$, which is a standard definition that $X$ is disconnected (and by the way also shows in the finite case that each component is both closed and open).
For the infinite case, the infinite union is not necessarily closed and this argument fails - is there a direct proof for the infinite case ?
Later clarification after many comments....
In other words, as per one of the comment below, given an infinite number of components can one construct a separation (disconnection) as one can in the finite case ?
A "direct proof" is clear: if $X$ is connected it has one unique component (where a component is defined as a maximal connected subset: $C$ is a component of $X$ when $C$ is non-empty, connected and $C \subsetneq C' \subseteq X$ implies $C'$ is not connected, or equivalently: $C \subseteq C'\subseteq X$ and $C'$ connected implies $C'=C$). Because when $X$ is connected then $C=X$ fulfills the definition I gave trivially, and no proper subset can be a component as witnessed by $C'=X$ as well.
So if $X$ has more than one component (the number is irrelevant) we already know $X$ is not connected, as you say by contrapositive.
The other argument you gave for finitely many components is irrelevant and not needed anyway, so the infinite case is a "fake problem". The first paragraph is all you need. A contrapositive implication is no problem.