Prove $ \int_0^t 2X_s \ dX_s = X_t^2-X_0^2-\langle X, X\rangle_t $ WITHOUT Ito's formula

It sounds like you were almost there. Just use \begin{align}\tag{1} \int_0^tX_s\,dX_s&=\lim_{\max\limits_i|t_i-t_{i-1}|\to 0}\sum_{i=1}^n X_{t_{i-1}}(X_{t_i}-X_{t_{i-1}}) \end{align} and \begin{align} \langle X,X\rangle_t&=\lim_{\max\limits_i|t_i-t_{i-1}|\to 0}\sum_{i=1}^n(X_{t_i}- X_{t_{i-1}})^2\\\tag{2} &=\lim_{\max\limits_i|t_i-t_{i-1}|\to 0}\sum_{i=1}^nX_{t_i}(X_{t_i}- X_{t_{i-1}})-\int_0^tX_s\,dX_s\,. \end{align} The sum in the first term of (2) is \begin{align} \sum_{i=1}^nX_{t_i}(X_{t_i}- X_{t_{i-1}})&=\sum_{i=1}^nX_{t_i}^2- X_{t_{i-1}}^2+X_{t_{i-1}}^2-X_{t_i}X_{t_{i-1}}\\&=X_t^2-X_0^2-\sum_{i=1}^nX_{t_{i-1}}(X_{t_i}- X_{t_{i-1}})\,. \end{align} By (1) this is $X_t^2-X_0^2-\int_0^tX_s\,dX_s$ in the limit.

Altogether we have

$$ \langle X,X\rangle_t=X_t^2-X_0^2-2\int_0^tX_s\,dX_s\,. $$