[proofplan]
The proof uses the conjugation symmetry of a complexified real vector bundle. Fiberwise complex conjugation gives a complex vector bundle isomorphism from $E_{\mathbb C}$ to its conjugate bundle $\overline{E_{\mathbb C}}$. Chern classes are invariant under complex vector bundle isomorphism, while the Chern classes of a conjugate complex vector bundle satisfy the sign rule $c_j(\overline F)=(-1)^j c_j(F)$. Applying this with $j=2k+1$ gives $c_j(E_{\mathbb C})=-c_j(E_{\mathbb C})$, and de Rham cohomology over $\mathbb C$ has no $2$-torsion.
[/proofplan]
[step:Construct the conjugation isomorphism from the complexification to its conjugate bundle]
Let $F:=E_{\mathbb C}$, regarded as a smooth complex vector bundle over $M$. For each $p\in M$, define the fiber map
\begin{align*}
\kappa_p:F_p&\to \overline F_p
\end{align*}
\begin{align*}
v\otimes z&\mapsto v\otimes \overline z
\end{align*}
for $v\in E_p$ and $z\in\mathbb C$. Here $\overline F$ denotes the conjugate complex vector bundle: it has the same underlying real vector bundle as $F$, but complex scalar multiplication is given by
\begin{align*}
\lambda\cdot_{\overline F} w:=\overline\lambda\, w
\end{align*}
for $\lambda\in\mathbb C$ and $w\in F_p$, where the multiplication on the right is the original scalar multiplication in $F_p$.
The map $\kappa_p$ is complex-linear as a map $F_p\to\overline F_p$. Indeed, for $\lambda\in\mathbb C$,
\begin{align*}
\kappa_p(\lambda(v\otimes z))=v\otimes \overline{\lambda z}=v\otimes \overline\lambda\,\overline z.
\end{align*}
By the definition of scalar multiplication in $\overline F_p$, this is exactly
\begin{align*}
\lambda\cdot_{\overline F}\kappa_p(v\otimes z).
\end{align*}
The inverse fiber map is again $v\otimes z\mapsto v\otimes \overline z$, so the maps $\kappa_p$ assemble to a smooth complex vector bundle isomorphism
\begin{align*}
\kappa:F&\to \overline F.
\end{align*}
[guided]
The point of this step is to make precise the statement that a complexified real bundle is naturally equal to its own conjugate. Let
\begin{align*}
F:=E_{\mathbb C}=E\otimes_{\mathbb R}\mathbb C.
\end{align*}
For every point $p\in M$, the fiber is
\begin{align*}
F_p=E_p\otimes_{\mathbb R}\mathbb C.
\end{align*}
Define a map on this fiber by conjugating only the complex coefficient:
\begin{align*}
\kappa_p:F_p&\to \overline F_p
\end{align*}
\begin{align*}
v\otimes z&\mapsto v\otimes \overline z.
\end{align*}
This formula is well-defined because complex conjugation is $\mathbb R$-linear on $\mathbb C$, so it induces an $\mathbb R$-[linear map](/page/Linear%20Map) on the [tensor product](/page/Tensor%20Product) over $\mathbb R$.
The only point that needs care is complex linearity. The target is not $F_p$ with its original complex structure; it is the conjugate [vector space](/page/Vector%20Space) $\overline F_p$. In $\overline F_p$, scalar multiplication is defined by
\begin{align*}
\lambda\cdot_{\overline F} w:=\overline\lambda\,w,
\end{align*}
where the multiplication on the right is the original scalar multiplication in $F_p$. Thus, for $\lambda\in\mathbb C$, $v\in E_p$, and $z\in\mathbb C$, we compute
\begin{align*}
\kappa_p(\lambda(v\otimes z))=\kappa_p(v\otimes \lambda z)=v\otimes \overline{\lambda z}=v\otimes \overline\lambda\,\overline z.
\end{align*}
By the definition of the conjugate scalar multiplication, the last expression is
\begin{align*}
\lambda\cdot_{\overline F}(v\otimes \overline z)=\lambda\cdot_{\overline F}\kappa_p(v\otimes z).
\end{align*}
Therefore $\kappa_p$ is complex-linear as a map from $F_p$ to $\overline F_p$.
The inverse map is the same formula, since applying complex conjugation twice returns the original coefficient. Because this construction is fiberwise algebraic and agrees with the usual conjugation map in every local trivialization of $E$, the maps $\kappa_p$ vary smoothly with $p$. Hence they assemble to a smooth complex vector bundle isomorphism
\begin{align*}
\kappa:F\to \overline F.
\end{align*}
[/guided]
[/step]
[step:Compare the Chern classes of $E_{\mathbb C}$ and its conjugate bundle]
Since $\kappa:F\to\overline F$ is a complex vector bundle isomorphism covering $\operatorname{id}_M:M\to M$, the [naturality of Chern classes](/theorems/9771) under pullback, applied to the identity map and this bundle isomorphism, gives
\begin{align*}
c_j(F)=c_j(\overline F)
\end{align*}
in $H_{\mathrm{dR}}^{2j}(M;\mathbb C)$ for every integer $j\ge 0$.
[/step]
[step:Apply the conjugate bundle sign formula to odd Chern classes]
By [citetheorem:9778] applied to the smooth complex vector bundle $F\to M$, for every integer $j\ge 0$,
\begin{align*}
c_j(\overline F)=(-1)^j c_j(F)
\end{align*}
in $H_{\mathrm{dR}}^{2j}(M;\mathbb C)$. Combining this with the equality from the previous step gives
\begin{align*}
c_j(F)=(-1)^j c_j(F).
\end{align*}
Now fix an integer $k\ge 0$ and set
\begin{align*}
j:=2k+1.
\end{align*}
Then $j$ is odd, so $(-1)^j=-1$, and therefore
\begin{align*}
c_{2k+1}(F)=-c_{2k+1}(F).
\end{align*}
Equivalently,
\begin{align*}
2c_{2k+1}(F)=0.
\end{align*}
[/step]
[step:Use the vector space structure of de Rham cohomology to conclude vanishing]
The class $c_{2k+1}(F)$ lies in the complex vector space
\begin{align*}
H_{\mathrm{dR}}^{4k+2}(M;\mathbb C).
\end{align*}
Multiplication by $2\in\mathbb C$ is injective on every complex vector space. From
\begin{align*}
2c_{2k+1}(F)=0
\end{align*}
we therefore obtain
\begin{align*}
c_{2k+1}(F)=0.
\end{align*}
Since $F=E_{\mathbb C}$, this is precisely
\begin{align*}
c_{2k+1}(E_{\mathbb C})=0
\end{align*}
in $H_{\mathrm{dR}}^{4k+2}(M;\mathbb C)$ for every integer $k\ge 0$.
[/step]
