Subcover is Refinement of Cover

Theorem

Let $S$ be a set.

Let $\CC$ be a cover for $S$.

Let $\VV$ be a subcover of $\CC$.

Then $\VV$ is a refinement of $\CC$.

Let $T = \struct {X, \tau}$ be a topological space.

Let $\UU$ be an open cover for $S$.

Let $\VV$ be a subcover of $\UU$.

Then $\VV$ is an open refinement of $\UU$.

From definition of subcover:

$\VV \subseteq \CC$

That is, every element of $\VV$ is an element of $\CC$.

From definition of subset, every element of $\VV$ is the subset of some element of $\CC$.

This is precisely the definition of refinement.

$\blacksquare$