From Kresyzig's Functional Analysis:
Define the operator $T^*$ between two Hilbert spaces as the operator such that $\langle Tx, y \rangle = \langle x, T^* y \rangle$ where $T$ is an operator.
If order that $T^*$ be an operator for each $y$ which is supposed to belong to the domain $D(T^*)$ of $T^*$ the corresponding $y^* = T^*y$ must be unique. We claim this holds if and only if: $D(T)$ is dense in $H$.
Indeed, if $D(T)$ is not dense in $H$, then $\overline{D(T)} \ne H$, the orthogonal complement of $\overline{D(T)}$ in $H$ contains a nonzero $y_1$ and $y_1 \perp x$ for every $x \in D(T)$, that is, $\langle x, y_1 \rangle = 0$. But then we obtain $\langle x, y^* \rangle = \langle x, y^* \rangle + \langle x, y_1 \rangle = \langle x, y^* + y_1 \rangle$, which shows non-uniqueness.
Can someone explain how this shows non-uniqueness? How does $\langle x, y^* \rangle = \langle x, y^* \rangle + \langle x, y_1 \rangle = \langle x, y^* + y_1 \rangle$ imply that there's not a unique $y^*$ such that $T^* y = y^*$?
The element $y^{\ast} \in H$ is defined by the fact that $\forall x \in H: \langle Tx, y \rangle = \langle x, y^{\ast}\rangle$. So given that $D(T)$ is not dense, we want to show that there exist $y \neq z$ s.t. $\forall x \in H: \langle x, y\rangle = \langle x, z\rangle$. With $z:= y^{\ast} + y_1$ this is precisely what is given.
In other words we have $T^{\ast} y = y^{\ast} = y^{\ast} + y_1$, but $y^{\ast} \neq y^{\ast} + y_1$, which shows that for any $y \in H$ the corresponding $T^{\ast}y$ is not unique.
Note that this relies crucially on the fact that there exists a non-zero element $y_1 \in H$ s.t. $\langle x, y_1 \rangle = 0$.