In order to prove following theory, my textbook gives reasoning that $\left\langle Tu,w\right\rangle =0, \forall u,w$ implies $T=0$. But I am unable to understand where is the fact used in this proof, that it is strictly a complex inner product space (not real), because it is false for real inner product space.
Over $\mathbb{C}$, $T v$ is orthogonal to $v$ for all $v$ only for the $0$ operator.
We have $\left\langle Tu,w\right\rangle=\frac{\left\langle T(u+w),u+w\right\rangle-\left\langle T(u-w),u-w\right\rangle}{4}+i\frac{\left\langle T(u+iw),u+iw\right\rangle-\left\langle T(u-iw),u-iw\right\rangle}{4}.$
Each term on the RHS are of the form $\left\langle Tv,v\right\rangle$. Thus our hypothesis implies that $\left\langle Tu,w\right\rangle=0$, $\forall u,v\in V$. This implies that $T=0$ (taking $w=Tu$).
But real space is a subspace of a complex space. So, the proof should be valid for real space as well, which is not true?
In that proof, multiplication by $i$ is used. Therefore, the proof makes no sense in a real vector space.
On the other hand, your sentence “But real space is a subspace of a complex space.” is strange. Are you saying that every real vector space is a subspace of a complex vector space? What happens is that, for every real vector space $V$, there is a complex vector space such that $V$ is isomorphic to a real vector subspace of $W$. But that's a quite heavy approach for this problem.