Consider the topological space arising as the quotient $ X:=\mathbb{R}^3/\sim $ where we identify $ x\sim \pm x $.
Does there exist a locally trivial fibre bundle with base space $ \mathbb{R}^2 $ and total space $ X $?
The obvious projection $ \pi: X \to \mathbb{R}^2,\ (x,y,z)\mapsto (x,y) $ fails to give a fiber bundle since
$ \pi^{-1}(\{p\}) $ is homeomorphic to $\mathbb{R}$ if $ p\neq 0 $ and to $ [0,\infty) $ if $ p=0 $.
I know that $X$ is not a manifold (but $X\setminus\{0\}$ is!) and hence cannot be the total space of a locally trivial fiber bundle over $\mathbb{R}^2$ with manifold fibers.
But how to rule out any other fibers?
I am interested in this space in the context of metric geometry where it is an important example of a ''non-manifold'' but I seem to lack the topological tools to tackle this problem.
No such fiber bundle exists.
Let me use the notation $\mathcal O \in X$ for the image of the origin of $\mathbb R^3$ under the quotient map $\mathbb R^3 \mapsto \mathbb R^3 / \sim \,\, = X$.
Arguing by contradiction, suppose there exists a locally trivial fibration $\pi : X \to \mathbb R^2$. Let $p = \pi(\mathcal O)$, and let $F = \pi^{-1}(p)$. Choose a pair of disjoint open balls $B,C \subset \mathbb R^2$ such that $p \in B$; it follows that $\pi^{-1}(B)$ and $\pi^{-1}(C)$ are disjoint open subsets of $X$ and that $\mathcal O \in \pi^{-1}(B)$.
As with any two locally trivial fiber bundles over open balls in $\mathbb R^2$, the subspaces $\pi^{-1}(B)$, $\pi^{-1}(C)$ are homeomorphic to each other. In fact both are homeomorphic to the product of an open ball in $\mathbb R^2$ times the fiber $F$.
The rest is algebraic topology, and is independent of $F$. This is the key point where arbitrary fibers $F$ are ruled out, as you asked.
Choosing a homeomorphism $h : \pi^{-1}(B) \to \pi^{-1}(C)$, there is an induced homeomorphism of pairs $$h : (\pi^{-1}(B),\pi^{-1}(B) - \mathcal O) \to (\pi^{-1}(C),\pi^{-1}(C) - h(\mathcal O)) $$ and so $h$ induces isomorphisms of the various homotopy functors of pairs, such as relative homology groups.
However, a direct calculation of relative $H_3$ with $\mathbb Z$ coefficients shows that these two pairs are not homeomorphic. I suspect that you might be familiar with this calculation, because you know that $X$ is not a manifold.
Since $X$, and therefore $\pi^{-1}(C)$, is indeed a 3-manifold at the point $h(\mathcal O)$, we get $$(1) \qquad H_3(\pi^{-1}(C),\pi^{-1}(C) - h(\mathcal O)) \approx H_2(S^2) \approx \mathbb Z $$ However, at the point $\mathcal O$ itself, $X$ is locally homeomorphic to a cone on $\mathbb R P^2$, and therefore $$(2) \qquad H_3(\pi^{-1}(B),\pi^{-1}(B) - \mathcal O) \approx H_2(\mathbb R P^2) \approx \mathbb Z/2\mathbb Z $$ Just as a reminder, both of these isomorphisms follow by an argument using excision and long exact homology sequences. First use excision to replace the spaces $\pi^{-1}(C)$ and $\pi^{-1}(B)$ by particularly nice neighborhoods of $h(\mathcal O)$ and $\mathcal O$, respectively: an open $3$-ball around $h(\mathcal O)$, which is of course a cone on $S^2$; and a cone on $\mathbb RP^2$ around $\mathcal O$. Then apply the long exact homology sequence to the resulting pairs. Each of those two neighborhoods is contractible, and therefore the terms that occur in the long exact sequences just before and after each pair of terms shown (in each of (1) and (2)) are trivial.