[proofplan]
The assertion is local on $X$, so we compute in an arbitrary holomorphic frame of $E$. In such a frame the Chern connection has connection matrix $A = H^{-1}\partial H$, where $H$ is the Hermitian metric matrix; in particular $A$ has type $(1,0)$. The curvature matrix is $dA + A \wedge A$, so the only non-immediate point is the vanishing of its $(2,0)$ component. This follows by differentiating the identity $H^{-1}H = I$ and using $\partial^2 = 0$, leaving only the $(1,1)$ component.
[/proofplan]
[step:Compute the Chern connection in a holomorphic frame]
Let $p \in X$. Choose an open neighbourhood $U \subset X$ of $p$ over which $E$ admits a holomorphic frame $e = (e_1,\dots,e_r)$, where $r = \operatorname{rank}E$. Let
\begin{align*}
H: U \to \operatorname{Herm}_r^{+}(\mathbb{C})
\end{align*}
be the Hermitian metric matrix in this frame, defined by $H_{\alpha\beta} = h(e_\alpha,e_\beta)$ for $1 \leq \alpha,\beta \leq r$.
In the frame $e$, a smooth section of $E|_U$ is represented by a smooth map
\begin{align*}
s: U \to \mathbb{C}^r.
\end{align*}
Here $\bar{\partial}_E:A^0(U;E|_U)\to A^{0,1}(U;E|_U)$ denotes the local Dolbeault operator induced by the holomorphic structure. Since $D^{0,1}=\bar{\partial}_E$ and the frame $e$ is holomorphic, the local connection matrix has no $(0,1)$ part. Thus the Chern connection has the local expression
\begin{align*}
D s = d s + A s,
\end{align*}
where the connection matrix satisfies
\begin{align*}
A \in A^{1,0}(U;\operatorname{End}\mathbb{C}^r).
\end{align*}
We use the column-vector convention: $s^*Ht$ denotes the Hermitian pairing $h(s,t)$ in the frame $e$, and the connection acts on local column vectors by $Ds=ds+As$. Metric compatibility says that for smooth sections $s,t:U\to\mathbb{C}^r$ represented in this frame,
\begin{align*}
d(s^*Ht) = (D s)^*Ht + s^*H(D t).
\end{align*}
Comparing the $(1,0)$ part under this convention and using arbitrariness of $s$ and $t$ gives
\begin{align*}
\partial H = H A.
\end{align*}
Multiplying on the left by $H^{-1}$ gives
\begin{align*}
A = H^{-1}\partial H.
\end{align*}
Here $\partial H$ denotes the matrix whose $(\alpha,\beta)$ entry is the $(1,0)$-form $\partial H_{\alpha\beta}$. Since $H^{-1}$ is a matrix of smooth functions and $\partial H$ has type $(1,0)$ entrywise, this formula is consistent with $A$ having type $(1,0)$.
[/step]
[step:Write the curvature matrix and isolate its type components]
In the same frame, the curvature $F_D = D^2$ is represented by the matrix-valued two-form
\begin{align*}
\Omega = dA + A \wedge A \in A^2(U;\operatorname{End}\mathbb{C}^r).
\end{align*}
Since $A$ has type $(1,0)$, its [exterior derivative](/theorems/1525) decomposes as
\begin{align*}
dA = \partial A + \bar{\partial}A,
\end{align*}
with $\partial A$ of type $(2,0)$ and $\bar{\partial}A$ of type $(1,1)$. Also $A \wedge A$ has type $(2,0)$ because it is the wedge product of two matrix-valued $(1,0)$-forms. Hence the type decomposition of $\Omega$ is
\begin{align*}
\Omega^{2,0} = \partial A + A \wedge A.
\end{align*}
The $(0,2)$ component is zero because the local connection matrix has no $(0,1)$ part in a holomorphic frame, equivalently because $D^{0,1}=\bar{\partial}_E$ and the holomorphic structure is integrable, so $\bar{\partial}_E^2=0$.
[/step]
[step:Show that the $(2,0)$ component vanishes]
We compute $\partial A + A \wedge A$ using $A = H^{-1}\partial H$. Since $H^{-1}H = I_r$, applying $\partial$ entrywise gives
\begin{align*}
\partial(H^{-1})H + H^{-1}\partial H = 0.
\end{align*}
Multiplying this identity on the right by $H^{-1}$ gives
\begin{align*}
\partial(H^{-1}) = -H^{-1}(\partial H)H^{-1}.
\end{align*}
Now apply the Leibniz rule for $\partial$ to $A = H^{-1}\partial H$. Since $\partial^2H=0$ entrywise, we obtain
\begin{align*}
\partial A = \partial(H^{-1}) \wedge \partial H.
\end{align*}
Substituting the formula for $\partial(H^{-1})$ gives
\begin{align*}
\partial A = -H^{-1}(\partial H)H^{-1}\wedge \partial H.
\end{align*}
By matrix multiplication of matrix-valued forms, the right-hand side is exactly $-A\wedge A$. Therefore
\begin{align*}
\partial A + A\wedge A = 0.
\end{align*}
[guided]
The only component that requires computation is the $(2,0)$ part. The connection matrix is
\begin{align*}
A = H^{-1}\partial H,
\end{align*}
where $H:U\to \operatorname{Herm}_r^{+}(\mathbb{C})$ is the Hermitian metric matrix. Since $A$ is a matrix-valued $(1,0)$-form, the possible $(2,0)$ curvature contribution is
\begin{align*}
\Omega^{2,0} = \partial A + A\wedge A.
\end{align*}
We now prove that this expression is zero. The key identity comes from differentiating the inverse relation
\begin{align*}
H^{-1}H = I_r.
\end{align*}
Applying $\partial$ to both sides and using the Leibniz rule gives
\begin{align*}
\partial(H^{-1})H + H^{-1}\partial H = 0.
\end{align*}
Multiplying on the right by $H^{-1}$ yields
\begin{align*}
\partial(H^{-1}) = -H^{-1}(\partial H)H^{-1}.
\end{align*}
Now differentiate $A = H^{-1}\partial H$. The operator $\partial$ satisfies the graded Leibniz rule, and $\partial^2H=0$ entrywise because $\partial^2=0$ on scalar-valued forms. Thus
\begin{align*}
\partial A = \partial(H^{-1})\wedge \partial H.
\end{align*}
Substituting the inverse-derivative formula gives
\begin{align*}
\partial A = -H^{-1}(\partial H)H^{-1}\wedge \partial H.
\end{align*}
But $A\wedge A$ is precisely the matrix product wedge
\begin{align*}
A\wedge A = H^{-1}(\partial H)H^{-1}\wedge \partial H.
\end{align*}
Therefore
\begin{align*}
\partial A = -A\wedge A.
\end{align*}
Hence
\begin{align*}
\partial A + A\wedge A = 0.
\end{align*}
This proves that the $(2,0)$ component of the curvature vanishes.
[/guided]
[/step]
[step:Conclude that the curvature has pure type $(1,1)$]
On the neighbourhood $U$, the curvature matrix satisfies
\begin{align*}
\Omega = \bar{\partial}A.
\end{align*}
Thus $\Omega$ has type $(1,1)$ on $U$. Since $p \in X$ was arbitrary and type decomposition of $\operatorname{End}E$-valued forms is local, the global curvature form $F_D$ has no $(2,0)$ or $(0,2)$ component. Therefore
\begin{align*}
F_D \in A^{1,1}(X;\operatorname{End}E).
\end{align*}
This proves the theorem.
[/step]
