[proofplan]
Over $\mathbb{F}_2$, every free rank-one module has a unique generator. Since $H^d(E_x, E_x^\#; \mathbb{F}_2) \cong \mathbb{F}_2$ for each fiber, there is exactly one choice of generator $\varepsilon_x$ at each point $x \in X$. The local compatibility condition for an $R$-orientation is therefore automatically satisfied: any two trivializations yield the same generator because $\mathbb{F}_2$ has no unit other than $1$.
[/proofplan]
[step:Identify the unique generator in each fiber]
For each $x \in X$, the fiber $E_x$ is a $d$-dimensional real vector space. The local cohomology of the fiber pair is
\begin{align*}
H^d(E_x, E_x^\#; \mathbb{F}_2) \cong H^d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; \mathbb{F}_2) \cong \mathbb{F}_2,
\end{align*}
where $E_x^\# = E_x \setminus \{0\}$ denotes the punctured fiber. Since $\mathbb{F}_2$ is a field with exactly two elements $\{0, 1\}$, the module $\mathbb{F}_2$ has a unique nonzero element, namely $1$. Therefore there is exactly one generator $\varepsilon_x \in H^d(E_x, E_x^\#; \mathbb{F}_2)$, and no choice is involved.
[guided]
Recall that an $R$-orientation of a $d$-dimensional vector bundle $\pi: E \to X$ consists of a family of generators $\{\varepsilon_x\}_{x \in X}$, where each $\varepsilon_x$ generates $H^d(E_x, E_x^\#; R) \cong R$ as an $R$-module, subject to a local compatibility condition: for each $x \in X$ there is a neighbourhood $U$ and a class $\varepsilon_U \in H^d(\pi^{-1}(U), \pi^{-1}(U)^\#; R)$ restricting to $\varepsilon_y$ for every $y \in U$.
Over a general ring $R$, the module $R$ has two generators: $+1$ and $-1$ (the units of $R$). The obstruction to orientability arises when transition functions force a switch between these two generators around a loop. Over $\mathbb{F}_2$, however, $R = \mathbb{F}_2 = \{0, 1\}$ has a single unit: $1 = -1$ in $\mathbb{F}_2$. Therefore $H^d(E_x, E_x^\#; \mathbb{F}_2) \cong \mathbb{F}_2$ has exactly one generator, namely $1$. There is no ambiguity and no choice to be made at any fiber.
[/guided]
[/step]
[step:Verify the local compatibility condition]
Let $\{U_\alpha\}_{\alpha \in I}$ be a trivializing cover for $E$, with trivializations $\varphi_\alpha: E|_{U_\alpha} \xrightarrow{\sim} U_\alpha \times \mathbb{R}^d$. For each $\alpha$ and each $x \in U_\alpha$, the trivialization induces an isomorphism on fiber cohomology:
\begin{align*}
(\varphi_\alpha)_x^*: H^d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; \mathbb{F}_2) \xrightarrow{\sim} H^d(E_x, E_x^\#; \mathbb{F}_2).
\end{align*}
Since both sides are isomorphic to $\mathbb{F}_2$ and $(\varphi_\alpha)_x^*$ is an isomorphism, it sends the unique generator of the source to the unique generator $\varepsilon_x$ of the target. In particular, if $x \in U_\alpha \cap U_\beta$, the two trivializations $\varphi_\alpha$ and $\varphi_\beta$ produce the same generator $\varepsilon_x$: the transition map on cohomology is an automorphism of $\mathbb{F}_2$, which must be the identity since $\operatorname{Aut}_{\mathbb{F}_2}(\mathbb{F}_2) = \{1\}$.
The class $\varepsilon_{U_\alpha} \in H^d(E|_{U_\alpha}, E|_{U_\alpha}^\#; \mathbb{F}_2)$ obtained from the Künneth isomorphism for the trivial bundle $U_\alpha \times \mathbb{R}^d$ restricts to $\varepsilon_x$ for every $x \in U_\alpha$. This establishes the local compatibility condition.
[guided]
The local compatibility condition asks: can we choose the generators $\varepsilon_x$ so that they "vary continuously" — more precisely, so that they are restrictions of a single cohomology class over a neighbourhood?
For a trivial bundle $U \times \mathbb{R}^d$, the relative Künneth theorem provides a canonical class in $H^d(U \times \mathbb{R}^d, U \times (\mathbb{R}^d \setminus \{0\}); \mathbb{F}_2)$ that restricts to the unique generator at each fiber. So local compatibility holds on each trivializing open set $U_\alpha$.
What could go wrong on overlaps? If $x \in U_\alpha \cap U_\beta$, the two trivializations $\varphi_\alpha$ and $\varphi_\beta$ define $\varepsilon_x$ via two different isomorphisms. These differ by the action of the transition map $g_{\alpha\beta}(x): \mathbb{R}^d \to \mathbb{R}^d$ on $H^d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; \mathbb{F}_2)$. This action is multiplication by $\operatorname{sgn}(\det g_{\alpha\beta}(x)) \in \{+1, -1\}$. Over $\mathbb{Z}$, the sign could be $-1$, switching the generator — this is the orientation obstruction. Over $\mathbb{F}_2$, however, $+1 = -1$, so the action is always the identity. The two trivializations produce the same $\varepsilon_x$, and the local classes $\varepsilon_{U_\alpha}$ and $\varepsilon_{U_\beta}$ agree on the overlap.
Since the local compatibility condition is satisfied over each trivializing set and the choices are consistent on overlaps, $E$ is $\mathbb{F}_2$-orientable.
[/guided]
[/step]
[step:Conclude that $E$ is $\mathbb{F}_2$-orientable]
The family $\{\varepsilon_x\}_{x \in X}$, where $\varepsilon_x$ is the unique generator of $H^d(E_x, E_x^\#; \mathbb{F}_2) \cong \mathbb{F}_2$, satisfies the local compatibility condition established in the previous step. Therefore $\pi: E \to X$ is $\mathbb{F}_2$-orientable.
[/step]
