= Solution
The <Hilbert projection theorem> states that every vector $x$ has a unique decomposition $x=m+n$ relative to a closed <linear subspace> $M$, with $m\in M$ and $n\in M^\perp$. The vector $m$ is the unique nearest point in $M$.
The <Riesz representation theorem> states that every bounded complex-<linear functional> $F$ on a <Hilbert space> has a unique representation $F(x)=\langle x,y\rangle$, with $\|F\|=\|y\|$, using the convention of linearity in the first argument. If $F=0$, use $y=0$. Otherwise $M=\ker F$ is closed. Project a vector outside $M$ to obtain $z\in M^\perp$ with $F(z)\ne0$, and rescale it so $F(z)=1$. Then $x-F(x)z\in M$, so
$$
\langle x,z\rangle=F(x)\|z\|^2,\qquad y=z/\|z\|^2.
$$
This gives the representation. Uniqueness follows because a vector orthogonal to every $x$ is zero. The <Cauchy-Schwarz inequality> gives $\|F\|\leq\|y\|$, and evaluation at $y/\|y\|$ gives equality when $y\ne0$.
For the adjoint, fix $v$. The functional $u\mapsto\langle Tu,v\rangle$ is bounded with <norm> at most $\|T\|\|v\|$. By the <Riesz representation theorem>, there is a unique vector $T^*v$ satisfying
$$
\boxed{\langle u,T^*v\rangle=\langle Tu,v\rangle\quad(u\in V).}
$$
For scalars $a,b$, conjugate-linearity in the second argument shows that $T^*(av+bw)=aT^*v+bT^*w$, by uniqueness of the representing vector. The same <norm> estimate gives $\|T^*v\|\leq\|T\|\|v\|$. Thus $T^*$ is a unique <bounded linear operator>, the <adjoint operator>.
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