Proving that the map between Hilbert spaces is unique. 
Theorem: Let $T:H\rightarrow K$ be a bounded linear operator between Hilbert spaces. Then there exists a unique bounded linear operator $T^* :K\rightarrow H$ such that
$$\tag{1}
\forall h\in H,k\in K :\langle Th,k\rangle=\langle h,T^*k\rangle.
$$

The following proof of existence was given by by lecturer:

Suppose there is another linear operator, $S :K\rightarrow H$ such that
$$\tag{2}
\forall h\in H,k\in K :\langle Th,k\rangle=\langle h,Sk\rangle.
$$
Then
$$\tag{3}
\left\langle h, T^{*} k-S k\right\rangle=\left\langle h, T^{*} k\right\rangle-\langle h, S k\rangle=\langle T h, k\rangle-\langle T h, k\rangle=0 .
$$
Thus $T^{*} k=S k$ for all $k\in K$, and therefore $T^*=S$.

What confuses me is that in eq. $(2)$ he defined $S$ such that $\langle h,Sk\rangle=\langle h,T^*k\rangle$ for all $k\in K$, and I don't see why eq. $(3)$ makes it more obvious that $T^*=S$? What exactly is eq. $(3)$ showing that is not obvious from eq. $(1)$ and $(2)$?

This is using the fact that if a vector $v\in H$ satisfies $\langle h,v\rangle=0$ for all $h\in H$, then $v=0$.  (To prove this, just take $h=v$ and use the positive-definiteness of the inner product.)  Applying this with $v=T^*k-Sk$, equation (3) tells you that $T^*k-Sk=0$ so $T^*k=Sk$.
