Let $T$ be normal. By the spectral theorem, there is a unitary map $V : H \to L^2(\mu)$ for a suitable measure $\mu$ on a measure space $X$ such that we have
$$
T = V^\ast M_f V
$$
for some bounded function $f : X \to \Bbb{C}$, where $M_f : L^2(\mu) \to L^2 (\mu), g \mapsto f\cdot g$ is the multiplication operator which multiplies by $f$.
By the usual properties of the spectral calculus, we have $\varphi(T) = V^\ast M_{\varphi \circ f} V$ for every bounded measurable $\varphi : \Bbb{C} \to \Bbb{C}$. In particular, we have $|T|= V^\ast M_{|f|} V$.
Now, it is easy to see that there is a measurable $g : X \to \Bbb{C}$ with $|g(x)| = 1$ for all $x \in X$ and $|f(x)| \cdot g(x) = f(x)$. In fact, the function
$$
g : X \to S^1 \subset \Bbb{C}, x \mapsto \begin{cases} \frac{f(x)}{|f(x)|}, & f(x) \neq 0,\\
1, & f(x) = 0\end{cases}
$$
does the job.
Since $g$ has modulus one, the multiplication operator $M_g$ is unitary. Hence, so is $U = V^\ast M_g V$ and we have
$$
T = V^\ast M_{f} V = V^\ast M_g M_{|f|} V = V^\ast M_g VV^\ast M_{|f|} V = U\cdot |T|
$$
as desired.
EDIT: As I wrote in the comments, the unitary $U$ is not unique in general. For example for $T=0$, we have $|T|=0$, so that any unitary $U : H \to H$ will do the job.
I read through the exercise in Conway's book
(Second edition from 1990, Chapter II, ยง7, Exercise 16, Part c, it's on page 59),
and it deals with the polar decomposition of $\,T=U\!A\,$ for some bounded operator $T$ on $H$. The bounded operators $A$ and $U$ are uniquely determined by $T$ when asked to satisfy
- $A=\sqrt{T^*T}=|T| $
- $U$ is a partial isometry with $\,\ker U=\ker T\,$ and
$\,\operatorname{im} U =\overline{\,\operatorname{im}T}\,$.
Note that $\,U^*\!\!\:U\,$ is the projector onto $\,\overline{\operatorname{im}|T|}\,$ which implies $\,A=U^*\!\!\:U\!A=U^*T$.
Multiplying with $U$ from the left yields $\,UU^*T=T$.
Consequences for the polar decomposition of $\,T^*$
$$=(U\!A)^*=AU^*= U^*\big(U\!AU^*\big)$$
Thus $\,U|T|U^*=\,|T^*|\,$ and $\,T^*=U^*|T^*|\,$ is the (unique) polar decomposition of $\,T^*$.
In the sequel $T$ is not required to be compact. I would guess that mentioning that property is due to the early placement of the exercise in the book in which Conway proceeds from the special to the more general.
It is shown that $\:T=AU\iff T$ is normal.
"$\,\Longrightarrow\,$"
If $T=U\!A=AU\,$ then $T^*=AU^*=U^*\!A$, so the assumption tells us that $A$ also commutes with $U^*$. Hence
$$\,T^*= U^*\big(U\!AU^*\big)= U^*UU^*\!A = U^*\!A$$
is the polar decomposition of $\,T^*$. We have $\,UT^*=A\,$ by the above "Note that ..." applied to $\,T^*$, and conclude that $T\,T^*=A\,UT^*=A^2=T^*T$.
"$\,\Longleftarrow\,$"
$\,T=U\!A\,$ yields $\,T^*=AU^*$. Hence $\,UA^2U^* =T\,T^*=T^*T= A^2\,$ by normality. Taking the positive square-root (which is unique) gives
$\,U\!AU^* = A$, being equivalent to $\,UA = A\,U$ by the properties of $U$.
Best Answer
It is easy to see that $|T|^2$ is self-adjoint, so it has a square root which is self-adjoint. Now $|T|^2$ is also a positive operator, since $$<T^*Tx,x> = <Tx,Tx> = ||Tx||^2\geq0$$ for all $x\in H$. Which implies that the square root is unique, hence $|T|$ is self-adjoint. So we find that $T^* = |T|U^*$.
Take $x\in \operatorname{ker}T^*$. We know that $\operatorname{ker}T = \operatorname{ker}|T| = \{0\}$. Since $|T|U^*x = 0$, $x\in\operatorname{ker}U^*$. But since $U$ is unitary we get $$0 = <U^*x,U^*x> = <x,x> = ||x||^2$$ So $x= 0$.