Does every cover have an irredundant subcover?

As often happens when I write questions here, I realized something while writing. Here I actually realized the answer.

It is no.

Here is an open cover of $\mathbb{R}$ with no irredundant subcover:

$\Lambda=\mathbb{N}$.

$U_n = (-n,n)$.

Because $U_1\subset U_2\subset U_3\subset\dots$, given any pair of $U_n$'s, one is contained in the other. Therefore any subcover is redundant unless it consists of a single $U_n$. But no single $U_n$ forms a cover. Thus no subcover is irredundant.

As you’ve already discovered, the answer is no. In fact, you can’t even guarantee an irreducible open refinement. For $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k<n\}$, and let $\tau=\{U_n:n\in\Bbb N\}\cup\{\Bbb N\}$; then $\tau$ is a $T_0$ topology on $\Bbb N$, and $\tau\setminus\{\Bbb N\}$ is an open cover of $\Bbb N$ with no irreducible open refinement.

A well-known positive result is that every point-finite open cover of a space has an irreducible subcover. Thus, every open cover of a metacompact space has an irreducible open refinement. (A space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement. A family of sets is point-finite if each point of the space lies in only finitely many of the sets.)