How to calculate $\int_{z=0}^T \int_{s=0}^z dW(s)dW(z)$ We are given the following integral
$$
\int_{z=0}^T \int_{s=0}^z dW(s)dW(z).
$$
My approach was to discretize the integral, obtaining
$$
\lim_{n \rightarrow \infty} \sum_{i = 1}^{n-1}\left[\lim_{m \rightarrow \infty} \sum_{j = 1}^{m - 1} W(t_j) - W(t_{j-1}) \right](W(t_i) - W(t_{i-1}))
$$
where $t_i = i\frac{T}{n}$ and $t_j = i\frac{z}{m}$.
Then, by independence of Brownian Motion increments we can argue that the previous equals (not sure about this part)
$$
\lim_{n \rightarrow \infty} \sum_{i = 1}^{n-1}(W(t_i) - W(t_{i-1}))^2.
$$
for which I can show that it converges to $T$.
Am I solving this integral in the wrong way?

For example using the limit formula you presented you can see that
$$\int_{s=0}^zdW_s=W_z-W_0=W_z$$
where I normalized $W_0=0$. We then need to evaluate
$$\int_{z=0}^TW_zdW_z$$
You can again evaluate this using the limit formula. Alternatively apply Ito's lemma to $W_t^2$. You get
$$d(W_t^2)=2W_tdW_t+dt$$
i.e.
$$W_t^2=\int_{z=0}^T2W_tdW_t+T$$
and
$$\int_{z=0}^TW_tdW_t=0.5(W_T^2-T),$$
which is the final result.
