Suppose $X$ is a continuous local martingale (on some complete filtered probability space). Then I want to prove that $$ \int_0^t 2X_s \ dX_s = X_t^2-X_0^2-\langle X, X\rangle_t $$ where $\langle X , X \rangle$ denotes the quadratic variation process of $X$.
One way to go about this is using Ito's formula but I want to see if I can give a direct proof of this just using properties of the stochastic integral.
What I have thought about so far is that by definition of the quadratic variation process the above yields that $$ \int_0^t 2X_s \ dX_s - X_0^2 $$ is a continuous local martingale. Maybe this could be used along with the fact that the quadratic variation process is unique to show the relation?
It almost also looks like that $(a+b)^2=a^2+b^2+2ab$ could play a role. Maybe this would lead to too many terms but some of them could be indistinguashable from 0 (by showing they were both a continuous local martingale and a finite variation process).
The textbook I have is "Brownian Motion, Martingales, and Stochastic Calculus" by Jean-Francois Le Gall so any result prior to the section on Ito's formula is available.
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
$$ \langle X,X\rangle_t=X_t^2-X_0^2-2\int_0^tX_s\,dX_s\,. $$