I'm assuming you are asking about the uniqueness of $U$ (at least that's what Rudin's exercise asks).
The result is that $U$ is unique if and only if $T$ has dense range.
Note that, since $U$ is a unitary, $$\tag{1}(\mbox{Im}\,T^*)^\perp=\ker T=\ker T^*=(\mbox{Im}\,T)^\perp.$$
Assume first that $U$ is unique. Let $P$ be the orthogonal projection onto $(\mbox{Im}\,T)^\perp$. Note that $U$ maps $\overline{\mbox{Im}\,T}$ into $\overline{\mbox{Im}\,T^*}=\overline{\mbox{Im}\,T}$. This means that $U(I-P)=(I-P)U(I-P)$. Multiplying by $P$ on the left, $PU(I-P)=0$. We can repeat the same reasoning for $T^*$, to get $PU^*(I-P)=0$. So $PU=PUP$, $PU^*=PU^*P$, and taking adjoint we get $UP=PUP=PU$. Now consider, for $\lambda\in\mathbb C$ with $|\lambda|=1$, $V=\lambda P+U(I-P)$; this is a unitary, since
$$
V^*V=(\overline\lambda P+(I-P)U^*)(\lambda P+U(I-P))=I+2\mbox{Re}\,\overline\lambda PU(I-P)=I,
$$
and
$$
VV^*=(\lambda P+U(I-P))(\overline\lambda P+(I-P)U^*)=P+U(I-P)U^*=P+(I-P)=I.
$$
Also, $VT=\lambda PT+U(I-P)T=0+(I-P)UT=(I-P)T^*=T^*$. By the uniqueness $U=V$, that is $\lambda P+U(I-P)=U=UP+U(I-P)$. So $\lambda P=UP$. As $\lambda$ was arbitrary, this shows that $P=0$. That is, $T$ has dense range.
Conversely, if $T$ has dense range, then $T$ and $T^*$ are injective, by $(1)$. Given $y\in\mbox{Im}\,T$, there exists $x$ such that $y=Tx$. This $x$ is unique by the injectivity of $T$. Then $Uy=UTx=T^*x$, so $U$ is defined uniquely on a dense subspace, and so it is defined uniquely.
A square root for an operator is not unique because square roots of complex numbers are not unique. If $A$ is a diagonal matrix on $\mathbb{C}^{n}$, then there are $2^{n}$ possible square roots for $A$ that you can spot right away. It's worse for a general Hilbert space.
If $N$ is bounded and normal on a Hilbert space, then the Spectral Theorem for $N$ gives a Borel spectral measure $E$ for which
$$
N = \int \lambda dE(\lambda).
$$
If $\sqrt{\lambda}$ is some branch of the square root, then you can define $\sqrt{N}$ by
$$
\sqrt{N} = \int \sqrt{\lambda} dE(\lambda).
$$
By the Borel functional calculus, $\sqrt{N}^{2}=\int\lambda dE(\lambda) = N$. By the way, you can't get this type of thing using the $C^{\star}$ algebra continuous functional calculus because $\sqrt{\lambda}$ cannot be assumed to be continuous on the spectrum of $N$. It's not that the result fails; it's that the technique of continuous functional calculus fails.
Check what assumptions you have for your version of the spectral theorem, especially concerning separability.
Best Answer
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.