For your first case:
Since
$U_n(x)
=\frac{\sin((n+1)t)}{\sin(t)}
$
where
$x = \cos(t)
$,
$\begin{array}\\
U_n(x)+U_{n-1}(x)
&=\frac{\sin((n+1)t)}{\sin(t)}+\frac{\sin(nt)}{\sin(t)}\\
&=\frac{\sin((n+1)t)+\sin(nt)}{\sin(t)}\\
&=\frac{2\sin((n+1/2)t)\cos(t/2)}{\sin(t)}\\
\end{array}
$
and this is zero when
$t(n+1/2)
=k\pi
$
for some integer $k$,
or
$t
=\frac{k\pi}{n+1/2}
$
for
$1 \le k \le n$.
This gives $n$ real roots,
and that is all
since
$U_n(x)+U_{n-1}(x)$
is of degree $n$.
Note that $U_n(x)=\frac {\sin (n+1)\theta}{\sin \theta}, x=\cos \theta$, so in particular the degree of $U_n$ is $n$ and its leading coefficient is $2^n$ since $\sin ((n+1)\theta)+\sin ((n-1)\theta)=2\cos \theta \sin (n\theta)$, so $U_n(x)=2xU_{n-1}(x)-U_{n-2}(x), n \ge 2, U_0(x)=1, U_1(x)=2x$
Let $s(x)=\frac{x}{|x|}, x \ne 0, s(0)=0$ be the sign function for real $x$. We note that $f(x)s(f(x))=|f(x)| \ge f(x)s(g(x))$ for any pair of real functions $f,g$ as $|f(x)| \ge \pm f(x)$.
Lemma: $\int_{-1}^1U_k(x)s(U_n(x))dx=0, k=0,..,n-1$
Assuming the lemma is proved, it then implies (if $p$ has real coefficients, while otherwise, we apply it to $\Re p$ which is a monic polynomial of the same degree with real coefficients and $|p(x)| \ge |\Re p(x)|$) that
$\int_{-1}^1(p(x)-2^{-n}U_n(x))s(U_n(x))=0$ since $p(x)-2^{-n}U_n(x)$ has degree at most $n-1$ so it is a linear combination of $U_0, U_1,..U_{n-1}$
But now as $|p(x)| =p(x)s(p(x)) \ge p(x)s(U_n(x))$ as noted, we integrate and get:
$\int_{-1}^1|p(x)|dx \ge \int_{-1}^1p(x)s(U_n(x))dx=\int_{-1}^12^{-n}U_n(x)s(U_n(x))dx=2^{-n}\int_{-1}^1|U_n(x)|dx$ so we only need to prove the lemma and compute $\int_{-1}^1|U_n(x)|dx$ to finish
First the Lemma:
Remembering the definition and using $x=\cos \theta, 0 \le \theta \le \pi, \sin \theta \ge 0$:
$\int_{-1}^1U_k(x)s(U_n(x))dx=\int_0^{\pi}(\sin (k+1)\theta)s(\sin (n+1)\theta)d\theta$ and since the integrand is even, it is enough to prove $\int_{-\pi}^{\pi}(\sin (k+1)\theta)s(\sin (n+1)\theta)d\theta=0$ as that is twice our integral.
We use the usual Fourier trick and notice that if we let
$I_k=\int_{-\pi}^{\pi}e^{ik\theta}s(\sin (n+1)\theta)d\theta=\int_{-\pi+a}^{\pi+a}e^{ik(\theta+a)}s(\sin (n+1)(\theta+a))d\theta=\int_{-\pi}^{\pi}e^{ik(\theta+a)}s(\sin (n+1)(\theta+a))d\theta =e^{ika}\int_{-\pi}^{\pi}e^{ik\theta}s(\sin (n+1)(\theta+a))d\theta$
with first equality by substitution, second by periodicity.
But now for $a=\frac{\pi}{n+1}$ we clearly have $s(\sin (n+1)(\theta+a))=-s(\sin (n+1)\theta)$, so $I_k=-e^{i\frac{k\pi}{n+1}}I_k$, or $I_k=0, 1 \le k \le n$ and the lemma follows by noticing that our original integral is the imaginary part of $I_k$
For the last part:
$\int_{-1}^1|U_n(x)|dx=\int_0^{\pi}|\sin (n+1)\theta|d\theta=\frac{1}{n+1}\int_0^{(n+1)\pi}|\sin \theta|d\theta=\int_0^{\pi}|\sin \theta|d\theta=2$ so we are done (again we use substitution, periodicity and then $|\sin \theta|=\sin \theta, 0 \le \theta \le \pi$)
Note that the proof shows that we can have equality iff $p=2^{-n}U_n$
Best Answer
According to this article in Wikipedia, the Chebyshev polynomials of the second kind are defined by the recurrence relation \begin{align} U_0(x)&=1, \\ U_1(x)&=2x, \\ U_{n+1}(x)&=2x\,U_n(x)-U_{n-1}(x). \tag{1} \end{align} If $x=\cos\theta$, one can show that $U_n(x)=\frac{\sin(n+1)\theta}{\sin\theta}$ satisfies $(1)$. On the other hand, one may use the recurrence relation $(1)$ to define $U_n(z)$ for any $z\in\mathbb C$.