Let $M$ be a smooth manifold of dimension $n$, and $\mathcal{U}$ any open cover of $M$, in particular $\mathcal{U}$ is not necessarily good or finite. Consider then the geometric realization of the nerve of $\mathcal{U}$. Is this homotopic/weakly equivalent to a space of dimension $\leq n$ (e.g. covering dimension)?
As I understand it, the nerve theorem tells me that if my open cover is a good one, this geometric realization will automatically be weakly equivalent to my original space $M$. This clearly doesn't need to happen if my open cover is badly chosen, e.g. if $\mathcal{U} = \{M\}$ I will just get a zero-dimensional point. But what I am hoping for is that even an arbitrarily bad cover will still give me something which is "essentially" at most $n$-dimensional.
Feels somewhat intuitively true, but I am not quite sure where I would start with a proof or what results might be helpful. Any ideas?
No. For instance, let $M=\{0,1,2\}$ (a $0$-dimensional manifold) and consider the open cover consisting of $\{0,1\},\{0,2\},$ and $\{1,2\},$ whose nerve is $S^1$.
More generally, let $X$ be any simplicial complex, and let $M$ be the set of faces of $X$, considered as a discrete space. For each $x\in X$, let $U_x=\{m\in M:x\in m\}$. Then $(U_x)_{x\in X}$ is an open cover of $M$, and it is easy to see that its nerve is exactly $X$.
Note that similar statements also hold in higher dimensions: if $M$ is any nonempty positive-dimensional manifold and $X$ is any finite connected simplicial complex, then there is an open cover of $M$ whose nerve is $X$. To prove this, let $f:M\to |X|$ be any continuous surjection (which exists by the Hahn-Mazurkiewicz theorem, for instance). Now take the open cover of $|X|$ by open stars of vertices, whose nerve is $X$, and pull it back to an open cover of $M$ via $f$. Since $f$ is surjective, this pulled back open cover has the same nerve.
(The connectedness assumption cannot be dropped, though, since the nerve of any open cover of a connected space is connected.)