As mentioned in the question, the inequality
$$
\begin{align}\int(f+g)\\,d\nu\ge\int f\\,d\nu+\int g\\,d\nu&&{\rm(1)}\end{align}
$$
follows easily from the definition of the integral $\int f\\,d\nu$ (as the supremum of the integrals of nonnegative simple functions bounded by $f$). So, I'll just show the reverse inequality which will establish additivity of the integral.

Choose any nonnegative simple function $h\le f+g$. We need to show that $\int f\\,d\nu+\int g\\,d\nu\ge\int h\\,d\nu$. However, we can write $h=\sum_{k=1}^nc_k1_{A_k}$ for $c_k\in(0,\infty)$ and pairwise disjoint $A_k\in\mathcal{A}$. As long as it can be shown that $\int_{A_k}f\\,d\nu+\int_{A_k}g\\,d\nu\ge c_k\nu(A_k)$, then the required inequality will follow by summing over $k$ and applying (1). So, replacing $f,g$ by $1_{A_k}f,1_{A_k}g$ respectively (for a fixed $k$), we reduce to the case where $n=1$. Dividing through by $c_k$ reduces to $c_k=1$.

So, we have reduced to the situation with $f+g\ge1_A$ and just need to show that $\int f\\,d\nu+\int g\\,d\nu\ge\nu(A)$. Without loss of generality (capping $f,g$ by 1 if necessary), we further reduce to the case with $0\le f,g\le1$. Then, for each positive integer $N$, consider the simple functions
$$
\begin{align}
f_N&=\sum_{j=0}^{\lfloor N\rfloor}1_{f^{-1}((j/N,(j+1)/N])}\frac jN\le f,\\\\
g_N&=\sum_{j=0}^{\lfloor N\rfloor}1_{g^{-1}((j/N,(j+1)/N])}\frac jN\le g.
\end{align}
$$
We have $f_N+g_N\ge(1-\frac2N)1_A$. So, using additivity for simple functions
$$
\int f\\,d\nu+\int g\\,d\nu\ge\int f_N\\,d\nu+\int g_N\\,d\nu\ge\left(1-\frac2N\right)\nu(A).
$$
Letting $N$ increase to infinity gives the required inequality.