W.l.o.g. we may assume that $A$ is an isomorphism. The complexification $A\in\Hom(E^\C)$ is skew-symmetric and thus we have an orthogonal decomposition $$ \E^\C=\bigoplus_j\ker(A-ia_j)\oplus\ker(A+ia_j)=\bigoplus_j\ker(A^2+a_j^2) $$ where the sum is taken over all pairewise distinct $a_j^2 > 0$, i.e. $a_1^2,\ldots,a_m^2$ are all pairewise distinct eigen-values of $-A^2$. We consider each space $\ker(A^2+a_j^2)\sbe E$ individually; $x+i\otimes y\in\ker(A-i)\sbe\E^\C$ iff $Ax=-a_jy$ and $Ay=a_jx$, in particular $x_j,y_j\in\ker(A^2+a_j^2)\sbe E$ and since $A$ is skew symmetric: $$ \la x,y\ra=\la Ay,y\ra/a_j=0, $$ i.e. $x,y$ are orthogonal. The same argument applied to the space orthogonal to $x,y$ gives another pair of orthogonal vectors $u,v$ satisfying $Au=-a_jv$ and $Av=a_ju$. Continuing this way we get an orthonormal basis $x_1,\ldots,x_l,y_1,\ldots,y_l$ of $\ker(A^2+a_j^2)$ such that: $Ax_k=-a_jy_k$ and $Ay_k=a_jx_k$.
We also see that for $a_j\neq0$ the spaces $\ker(A^2+a_j^2)$ are of even dimension and we may obviuosly choose $a_j > 0$.