An infinitesimal $SO(N)$ transformation matrix can be written :
$$R_{ij} = \delta_{ij}+\theta_{ij}+O(\theta^2)$$
Now it has to be shown that $\theta_{ij}$ is real and anti-symmetric.
I've started with the orthogonality condition like following:
$$R^TR=\boldsymbol 1$$ $$\implies (R^TR)_{ij}=\delta_{ij}$$ $$\implies \sum_kR^T_{ik}R_{jk}=\delta_{ij}$$ $$\implies \sum_kR_{ik}R_{jk}=\delta_{ij}$$ $$\implies \sum_kR_{ik}R_{ik}=\delta_{ii}$$ $$\implies \sum_jR_{ij}R_{ij}=1$$
Now i can stick my infinitesimal form of $R_{ij} $ into the above formula:
$$\sum_j(\delta_{ij}+\theta_{ij}+O(\theta^2))(\delta_{ij}+\theta_{ij}+O(\theta^2))=1$$ $$\implies \sum_j (\delta_{ij}\delta_{ij}+2\theta_{ij}\delta_{ij}+O(\theta^2))=1$$
As you can easily see my calculations are going nowhere.
First,
$$R^{T}_{ij}=\delta_{ij}+\theta_{ij}^{T}=\delta_{ij}+\theta_{ji}$$
Now in the last sum you wrote, you get a different result,
$$\sum_j(\delta_{ij}+\theta_{ij}+O(\theta^2))(\delta_{ij}+\theta_{ji}+O(\theta^2))=\sum_j(\delta_{ij}+\theta_{ji})(\delta_{ij}+\theta_{ij})=\sum_j(\delta_{ij}\delta_{ij}+\delta_{ij}(\theta_{ij}+\theta_{ji})+\theta_{ij}\theta_{ji})=1$$
This is true, if $\theta_{ij}=-\theta_{ji}$. To prove that $\theta_{ij}$ is real, evaluate this $\mathrm{Im}[(R^{T}R)_{ij}]$. You will find that $\mathrm{Im}[(R^{T}R)_{ij}]=0$.