[proofplan]
We regard $T$ as an operator whose range lies in the closed Hilbert subspace $\ker S \subset H_1$. The assumed estimate makes the functional $T^*v \mapsto (f,v)_{H_1}$ well-defined and bounded on the range of the adjoint of this restricted operator. Hahn-Banach extends this functional to all of $H_0$, and the [Riesz representation theorem](/theorems/221) represents it by an element $u \in H_0$ with norm at most $C^{1/2}$. The adjoint identity then shows that $u \in \operatorname{Dom}(T)$ and $Tu=f$.
[/proofplan]
[step:Restrict the target of $T$ to the closed subspace $\ker S$]
Since $S:\operatorname{Dom}(S)\subset H_1 \to H_2$ is closed, its kernel
\begin{align*}
K := \ker S = \{w \in \operatorname{Dom}(S): S w = 0\}
\end{align*}
is a closed linear subspace of $H_1$. Equipped with the inner product inherited from $H_1$, $K$ is therefore a [Hilbert space](/page/Hilbert%20Space).
Define the operator
\begin{align*}
\widetilde T:\operatorname{Dom}(\widetilde T)\subset H_0 &\to K \\
u &\mapsto Tu
\end{align*}
by setting $\operatorname{Dom}(\widetilde T):=\operatorname{Dom}(T)$. This is well-defined because $S T=0$ implies $Tu \in K$ for every $u \in \operatorname{Dom}(T)$.
The operator $\widetilde T$ is densely defined because $T$ is densely defined. It is closed as an operator into $K$: if $u_j \in \operatorname{Dom}(T)$, $u_j \to u$ in $H_0$, and $\widetilde T u_j \to w$ in $K$, then $\widetilde T u_j = T u_j \to w$ in $H_1$; since $T$ is closed, $u \in \operatorname{Dom}(T)$ and $Tu=w$, hence $u \in \operatorname{Dom}(\widetilde T)$ and $\widetilde T u=w$.
Finally, the adjoint $\widetilde T^*: \operatorname{Dom}(\widetilde T^*)\subset K \to H_0$ satisfies
\begin{align*}
\operatorname{Dom}(\widetilde T^*)=\operatorname{Dom}(T^*)\cap K,
\qquad
\widetilde T^* v = T^*v
\end{align*}
for every $v\in \operatorname{Dom}(T^*)\cap K$. Indeed, for $v\in K$, the map
\begin{align*}
u \mapsto (\widetilde T u,v)_K=(Tu,v)_{H_1}
\end{align*}
is bounded on $\operatorname{Dom}(T)\subset H_0$ exactly when $v\in\operatorname{Dom}(T^*)$, and then its representing vector is $T^*v$.
[guided]
The equation $Tu=f$ is meant inside $H_1$, but the assumption $S T=0$ says that every value of $T$ actually lies in $\ker S$. Since $f$ also satisfies $S f=0$, the natural target space for the equation is not all of $H_1$, but the closed Hilbert subspace
\begin{align*}
K := \ker S = \{w \in \operatorname{Dom}(S): S w = 0\}.
\end{align*}
We first verify that $K$ is closed. Let $(w_j)_{j=1}^{\infty}$ be a sequence in $K$ with $w_j \to w$ in $H_1$. Since $S w_j=0$ for every $j$ and $0\to 0$ in $H_2$, closedness of $S$ implies $w\in \operatorname{Dom}(S)$ and $S w=0$. Hence $w\in K$. Therefore $K$ is a closed linear subspace of the [Hilbert space](/page/Hilbert%20Space) $H_1$, and so $K$ is itself a [Hilbert space](/page/Hilbert%20Space) with the inherited inner product.
Now define
\begin{align*}
\widetilde T:\operatorname{Dom}(\widetilde T)\subset H_0 &\to K \\
u &\mapsto Tu
\end{align*}
with $\operatorname{Dom}(\widetilde T)=\operatorname{Dom}(T)$. This is well-defined because the hypothesis $S T=0$ means precisely that $Tu\in \operatorname{Dom}(S)$ and $S(Tu)=0$ for every $u\in\operatorname{Dom}(T)$; hence $Tu\in K$.
We also need $\widetilde T$ to be a closed densely defined operator, because we will use its adjoint. Density follows immediately from density of $\operatorname{Dom}(T)$ in $H_0$. For closedness, suppose $u_j\in\operatorname{Dom}(T)$, $u_j\to u$ in $H_0$, and $\widetilde T u_j\to w$ in $K$. Convergence in $K$ is the same as convergence in $H_1$ because $K$ has the inherited norm, so $T u_j\to w$ in $H_1$. Since $T$ is closed, $u\in\operatorname{Dom}(T)$ and $Tu=w$. Thus $u\in\operatorname{Dom}(\widetilde T)$ and $\widetilde T u=w$.
Finally we identify the adjoint of $\widetilde T$. For $v\in K$, the adjoint condition for $\widetilde T$ asks whether the functional
\begin{align*}
u \mapsto (\widetilde T u,v)_K
\end{align*}
is bounded on $\operatorname{Dom}(\widetilde T)$ with respect to the $H_0$ norm. Since the inner product on $K$ is inherited from $H_1$, this functional is
\begin{align*}
u \mapsto (Tu,v)_{H_1}.
\end{align*}
This is exactly the condition that $v\in\operatorname{Dom}(T^*)$, and in that case the representing vector is $T^*v$. Therefore
\begin{align*}
\operatorname{Dom}(\widetilde T^*)=\operatorname{Dom}(T^*)\cap K,
\qquad
\widetilde T^*v=T^*v.
\end{align*}
[/guided]
[/step]
[step:Define a bounded functional on the range of $\widetilde T^*$]
Define a linear functional
\begin{align*}
\Lambda_0:\operatorname{Range}(\widetilde T^*)\subset H_0 &\to \mathbb C \\
\widetilde T^*v &\mapsto (f,v)_K
\end{align*}
for $v\in\operatorname{Dom}(\widetilde T^*)$.
This definition is independent of the choice of representative. If $\widetilde T^*v_1=\widetilde T^*v_2$, then $\widetilde T^*(v_1-v_2)=0$. Since $v_1-v_2\in\operatorname{Dom}(\widetilde T^*)=\operatorname{Dom}(T^*)\cap K$, the assumed estimate gives
\begin{align*}
|(f,v_1-v_2)_{H_1}|^2
\leq C\|T^*(v_1-v_2)\|_{H_0}^2
=0.
\end{align*}
Hence $(f,v_1)_{H_1}=(f,v_2)_{H_1}$. Since $f,v_1,v_2\in K$ and the inner product on $K$ is inherited from $H_1$, this proves well-definedness.
The same estimate gives boundedness. For every $v\in\operatorname{Dom}(\widetilde T^*)$,
\begin{align*}
|\Lambda_0(\widetilde T^*v)|
= |(f,v)_K|
= |(f,v)_{H_1}|
\leq C^{1/2}\|T^*v\|_{H_0}
= C^{1/2}\|\widetilde T^*v\|_{H_0}.
\end{align*}
Therefore $\Lambda_0$ is a bounded linear functional on $\operatorname{Range}(\widetilde T^*)$ with operator norm at most $C^{1/2}$.
[guided]
The goal is to build a vector $u\in H_0$ whose inner products against $\widetilde T^*v$ reproduce the values $(f,v)_K$. This motivates defining a functional first on the subspace $\operatorname{Range}(\widetilde T^*)\subset H_0$:
\begin{align*}
\Lambda_0:\operatorname{Range}(\widetilde T^*)\subset H_0 &\to \mathbb C \\
\widetilde T^*v &\mapsto (f,v)_K.
\end{align*}
There is one point to check before this is a legitimate definition: the same vector in $\operatorname{Range}(\widetilde T^*)$ may have two different representations. Suppose
\begin{align*}
\widetilde T^*v_1=\widetilde T^*v_2
\end{align*}
with $v_1,v_2\in\operatorname{Dom}(\widetilde T^*)$. Then $v_1-v_2\in\operatorname{Dom}(\widetilde T^*)=\operatorname{Dom}(T^*)\cap K$ and
\begin{align*}
T^*(v_1-v_2)=\widetilde T^*(v_1-v_2)=0.
\end{align*}
The assumed estimate applies to $v_1-v_2$, because this vector lies in $\operatorname{Dom}(T^*)\cap K=\operatorname{Dom}(T^*)\cap\ker S$. Thus
\begin{align*}
|(f,v_1-v_2)_{H_1}|^2
\leq C\|T^*(v_1-v_2)\|_{H_0}^2
=0.
\end{align*}
Therefore $(f,v_1-v_2)_{H_1}=0$, which is the same as
\begin{align*}
(f,v_1)_{H_1}=(f,v_2)_{H_1}.
\end{align*}
Since $f,v_1,v_2\in K$ and $K$ has the inherited inner product, this also gives $(f,v_1)_K=(f,v_2)_K$. Hence $\Lambda_0$ is well-defined.
The estimate also proves boundedness. For $v\in\operatorname{Dom}(\widetilde T^*)$,
\begin{align*}
|\Lambda_0(\widetilde T^*v)|
= |(f,v)_K|
= |(f,v)_{H_1}|
\leq C^{1/2}\|T^*v\|_{H_0}
= C^{1/2}\|\widetilde T^*v\|_{H_0}.
\end{align*}
Thus $\Lambda_0$ is continuous on the normed subspace $\operatorname{Range}(\widetilde T^*)$ and has norm at most $C^{1/2}$.
[/guided]
[/step]
[step:Extend the functional and represent it by a vector in $H_0$]
By the Hahn-Banach [Extension Theorem](/theorems/59) (citing a result not yet in the wiki: Hahn-Banach [Extension Theorem](/theorems/59)), $\Lambda_0$ extends to a bounded linear functional
\begin{align*}
\Lambda:H_0\to\mathbb C
\end{align*}
with
\begin{align*}
\|\Lambda\|_{H_0^*}=\|\Lambda_0\|\leq C^{1/2}.
\end{align*}
By the [Riesz Representation Theorem](/theorems/218) for Hilbert spaces (citing a result not yet in the wiki: [Riesz Representation](/theorems/67) Theorem), there exists $u\in H_0$ such that
\begin{align*}
\Lambda(w)=(u,w)_{H_0}
\end{align*}
for every $w\in H_0$, and
\begin{align*}
\|u\|_{H_0}=\|\Lambda\|_{H_0^*}\leq C^{1/2}.
\end{align*}
Consequently,
\begin{align*}
\|u\|_{H_0}^2\leq C.
\end{align*}
[guided]
We have a bounded linear functional $\Lambda_0$ on the subspace $\operatorname{Range}(\widetilde T^*)$ of $H_0$. The Hahn-Banach [Extension Theorem](/theorems/59) says that a bounded linear functional on a subspace of a [normed vector space](/page/Normed%20Vector%20Space) extends to the whole space without increasing its norm. Applying that theorem to the normed space $H_0$, the subspace $\operatorname{Range}(\widetilde T^*)$, and the functional $\Lambda_0$, we obtain a bounded linear functional
\begin{align*}
\Lambda:H_0\to\mathbb C
\end{align*}
such that $\Lambda$ agrees with $\Lambda_0$ on $\operatorname{Range}(\widetilde T^*)$ and
\begin{align*}
\|\Lambda\|_{H_0^*}=\|\Lambda_0\|\leq C^{1/2}.
\end{align*}
This use of Hahn-Banach is the only place where the functional is extended beyond the range of $\widetilde T^*$.
Next we use the [Hilbert space](/page/Hilbert%20Space) structure. The [Riesz Representation Theorem](/theorems/221) for Hilbert spaces states that every bounded linear functional on a [Hilbert space](/page/Hilbert%20Space) is represented uniquely by inner product with a vector in that [Hilbert space](/page/Hilbert%20Space). Applying it to $\Lambda\in H_0^*$, there exists $u\in H_0$ such that
\begin{align*}
\Lambda(w)=(u,w)_{H_0}
\end{align*}
for every $w\in H_0$. Moreover the representing vector has the same norm as the functional:
\begin{align*}
\|u\|_{H_0}=\|\Lambda\|_{H_0^*}\leq C^{1/2}.
\end{align*}
Squaring this inequality gives the required estimate
\begin{align*}
\|u\|_{H_0}^2\leq C.
\end{align*}
[/guided]
[/step]
[step:Use the adjoint identity to prove $\widetilde T u=f$ in $K$]
For every $v\in\operatorname{Dom}(\widetilde T^*)$, the extension property of $\Lambda$ and the definition of $\Lambda_0$ give
\begin{align*}
(u,\widetilde T^*v)_{H_0}
=
\Lambda(\widetilde T^*v)
=
\Lambda_0(\widetilde T^*v)
=
(f,v)_K.
\end{align*}
Thus the map
\begin{align*}
v\mapsto (f,v)_K
\end{align*}
on $\operatorname{Dom}(\widetilde T^*)$ is represented by $u$ through the adjoint pairing. By the definition of the adjoint of the densely defined operator $\widetilde T^*$, this means $u\in\operatorname{Dom}((\widetilde T^*)^*)$ and
\begin{align*}
(\widetilde T^*)^*u=f.
\end{align*}
Since $\widetilde T$ is densely defined and closed, $(\widetilde T^*)^*=\widetilde T$. Hence $u\in\operatorname{Dom}(\widetilde T)$ and
\begin{align*}
\widetilde T u=f.
\end{align*}
[guided]
The vector $u$ was chosen so that $\Lambda(w)=(u,w)_{H_0}$ for all $w\in H_0$. In particular, for every $v\in\operatorname{Dom}(\widetilde T^*)$, we may substitute $w=\widetilde T^*v$. Since $\Lambda$ extends $\Lambda_0$, we obtain
\begin{align*}
(u,\widetilde T^*v)_{H_0}
=
\Lambda(\widetilde T^*v)
=
\Lambda_0(\widetilde T^*v)
=
(f,v)_K.
\end{align*}
This identity is exactly the adjoint criterion for $u$ to lie in the domain of $(\widetilde T^*)^*$. Indeed, by definition, $u\in\operatorname{Dom}((\widetilde T^*)^*)$ if there exists a vector $g\in K$ such that
\begin{align*}
(u,\widetilde T^*v)_{H_0}=(g,v)_K
\end{align*}
for every $v\in\operatorname{Dom}(\widetilde T^*)$. The identity above shows that this vector is $g=f$. Hence
\begin{align*}
u\in\operatorname{Dom}((\widetilde T^*)^*)
\qquad\text{and}\qquad
(\widetilde T^*)^*u=f.
\end{align*}
Because $\widetilde T$ is densely defined and closed, it equals its double adjoint:
\begin{align*}
(\widetilde T^*)^*=\widetilde T.
\end{align*}
Therefore $u\in\operatorname{Dom}(\widetilde T)$ and
\begin{align*}
\widetilde T u=f.
\end{align*}
[/guided]
[/step]
[step:Return from the restricted operator to the original equation]
Since $\operatorname{Dom}(\widetilde T)=\operatorname{Dom}(T)$ and $\widetilde T u=Tu$ for every $u\in\operatorname{Dom}(T)$, the identity $\widetilde T u=f$ gives
\begin{align*}
Tu=f
\end{align*}
in $H_1$. The norm estimate obtained above gives
\begin{align*}
\|u\|_{H_0}^2\leq C.
\end{align*}
Thus there exists $u\in\operatorname{Dom}(T)$ solving $Tu=f$ with the asserted bound.
[/step]
