We have that $M$ is a manifold, $f \in C^\infty(M)$ and $X,Y$ and $Z$ are smooth vector fields on $M$. We also know that $Y$ and $Z$ agree in a neighbourhood $p \in M$. I now have to show that $[X,Y]|_p = [X,Z]|_p$.
I have tried the following: for every $f \in C^\infty(M)$ we know that $[fX,Y] = f[X,Y] - Y(f)X$. Is it true that I may assume that $Y(f)X = Z(f)X$ when I evaluate in $p$? Because then, we know that
$f[X,Y]|_p - f[X,Z]|_p = [X,Y]|_p - [X,Z]|_p$ for all $f$ and hence $[X,Y]|_p = [X,Z]|_p$.
If $Y$ and $Z$ agree on a neighborhood of $p$, then on such neighborhood we have $Y(f) = Z(f)$ for all $f \in \mathcal{C}^\infty(M)$. Thus $X_p(Y(f)) = X_p(Z(f))$. Moreover, we have that $Y_p(X(f)) = Z_p(X(f))$, since in particular $Y_p=Z_p$. With this, we compute $$[X,Y]_p(f) = X_p(Y(f)) - Y_p(X(f)) = X_p(Z(f)) - Z_p(X(f)) = [X,Z]_p(f).$$