For $T: X \to Y$ with Hilbert $X,Y$, show that if $T^* Tx = T^*y$ then $x$ minimizes ${\lVert Tx - y \rVert}$
By hypothesis $T^* Tx = T^* y$, so $T^*(Tx - y) = 0$, so $Tx - y \in \mathop{\text{null}} T^*$. There is a proposition that says $\mathop{\text{null}} T^* = (\mathop{\text{im}} T)^\perp$ so $Tx - y \in \mathop{\text{null}} T^* = (\mathop{\text{im}} T)^\perp$. If $\mathop{\text{im}} T$ is closed then there is a proposition that says $Tx$ minimizes ${\lVert Tx-y \rVert}$ and completes the proposition but we can't make that assumption. I'm stuck on this proof when we can't assume that $\mathop{\text{im}} T$ is closed.
One more useful piece:
Let $M := \overline{\mathop{\text{im}} T} \subset Y$. For each $y \in Y$, a textbook proposition says that there exists a unique $y_* \in M$ that minimizes ${\lVert y - y_* \rVert}$ such that
\begin{align*} \forall m \in M, {\lVert y - y_* \rVert} \le {\lVert y - m \rVert} \\ \end{align*}
Another proposition says $y - y_* \in M^\perp$. Since $\mathop{\text{im}} T \subset M$ then $M^\perp \subset (\mathop{\text{im}} T)^\perp$ and another proposition says $\mathop{\text{nullspace}} T^* = (\mathop{\text{im}} T)^\perp$ so:
\begin{gather*} y - y_* \in M^\perp \subset (\mathop{\text{im}} T)^\perp = \mathop{\text{null}} T^* \\ T^*(y - y_*) = 0 \\ T^*y = T^* y_* \\ \end{gather*}
We can combine this with the hypothesis to get $T^* Tx = T^*y = T^* y_*$ such that $Tx - y_* \in \mathop{\text{null}} T^* = (\mathop{\text{im}} T)^{\perp}$
\begin{align*} \|Tz-y\|^2&=\|T(z-x)+Tx-y\|^2 \\ &=\|Tx-y\|^2+\|T(z-x)\|^2+2\langle T(z-x),Tx-y\rangle \\ &=\|Tx-y\|^2+\|T(z-x)\|^2 \\ &\geq \|Tx-y\|^2 \end{align*} for all $z$.