Lots of questions here. I think I've had a crack at all of them, but let me know if I've missed something, or if I've made a mistake somewhere. I might be light on detail in a lot of places; apologies in advance.
First: Why is proving the result for functions on $[0,1]$ sufficient?
Given that we have the result for functions on $[0,1]$, we can define the function $$g(x) = f(a + (b-a)x),$$ and apply the result to $g$, which is a function on $[0,1]$. This gives us a sequence of polynomials $\{ P_n \}$ on $[0,1]$. Given that
$$ f(x) = g\left(\frac{x-a}{b-a}\right), $$
and $P_n\left(\frac{x-a}{b-a}\right)$ is also polynomial, we have the desired sequence of polynomials approximating $f$.
Second: Why is $f$ uniformly continuous?
$f$ is a continuous function on a compact set, and is thus uniformly continuous on $[0,1]$. It is also constant outside of $[0,1]$, so there is nothing going on on the real line that "breaks" uniform continuity. Showing this rigorously entails a simple application of the definition of uniform continuity once you've established by the argument above that $f$ is uniformly continuous on $[0,1]$.
Alternatively, you can view this as a consequence of the pasting lemma for uniformly continuous functions.
Third: Is the fact that $Q_n \to 0$ uniformly on $\delta \le \lvert x \rvert \le 1$ needed in the proof?
You're right, this specific fact seems to play no role in the proof. That is, once we have the bound that $Q_n (x) \le \sqrt n \left( 1- \delta^2 \right)^n $ for $\delta \le \lvert x \rvert \le 1$, we have no need for the uniform convergence of $Q_n$. This is but natural, since the aforementioned bound is what implies uniform convergence.
Of course, one can still prove the theorem (in particular, the second to the last inequality) by dispensing with the specific bound and using the fact that $Q_n \to 0$ uniformly, but that seems to require extra notation, if not some work to set it up.
Fourth: How do you demonstrate that $ \int_0^1 f(t) Q_n (t -x) \, \mathrm dt $ is a polynomial in $x$?
Observe that
$$ \int_0^1 f(t) Q_n (t -x) \, \mathrm dt = \int_0^1 f(t) c_n \left(1 - (t-x)^2\right)^n \, \mathrm d t. $$
Standard manipulations (like, say, the binomial theorem) allow you to expand the expression in the integral. Any terms involving $t$ are integrated out, and all you're left with is a function of $x$ that is (hopefully) clearly a polynomial.
Finally: Should $\delta \in (0,1)$?
I think Rudin snuck this assumption in without making it clear. See, for example, $(50)$. Note also the condition that $\lvert y-x \rvert < \delta$ seems to suggest that $\delta \le 1$, since $x,y \in [0,1]$. In any case, you actually do need $\delta \in (0,1)$ for the last inequality to work for large $n$.
Say a function $f : [a, b] \to \mathbb{R}$, where $a < b$, "has the convergence property" if and only if there is a sequence of polynomials with real coefficients $\{P_i\}_{i \in \mathbb{N}}$ which converges uniformly to $f$ on $[a, b]$ and, furthermore, that $P_i = 0$ for all $i$ whenever $f = 0$ everywhere.
Rudin proves the following theorem:
Theorem 1: For all continuous $f : [0, 1] \to \mathbb{R}$ such that $f(0) = f(1) = 0$, $f$ has the convergence property.
It sounds like you have no questions about the proof of this theorem.
From this theorem, we can then prove the theorem
Theorem 2: For all continuous $f : [0, 1] \to \mathbb{R}$, where $a \leq b$, $f$ has the convergence property.
To prove theorem 2 from theorem 1, define $g(x) = f(x) + x(f(0) - f(1)) - f(0)$. Then $g : [0, 1] \to \mathbb{R}$ is continuous, and $g(0) = g(1) = 0$. So we can take a sequence of polynomials $\{P_i\}_{i \in \mathbb{N}}$ which uniformly converge to $g$ over $[0, 1]$. And furthermore, if $f$ was zero everywhere, then $g$ is zero everywhere, and hence all $P_i$ are zero everywhere.
Define $Q_i(x) = P_i(x) + f(0) + x(f(1) - f(0))$. Then we see that $\{Q_i\}_{i \in \mathbb{N}}$ converge uniformly to $g$. And if the $P_i$ are zero everywhere and $f(0) = f(1) = 0$, then the $Q_i$ are zero everywhere.
From here, we prove
Theorem 3: For all continuous $f : [a, b] \to \mathbb{R}$ where $a < b$, $f$ has the convergence property.
To prove this theorem, define $g(x) = f(x(b - a) + a)$, $g : [0, 1] \to \mathbb{R}$. Now $g$ is continuous and thus has the convergence property by Theorem 2. Take a sequence of polynomials $\{P_i\}_{i \in \mathbb{N}}$ which uniformly converge to $g$ over $[0, 1]$. Then define $Q_i(x) = P_i(\frac{x - a}{b - a}))$. We can show that $\{Q_i\}_{i \in \mathbb{N}}$ converges to $f$ over $[a, b]$. And if $f$ is zero everywhere, then $g$ is zero everywhere, so all $P_i$ are zero, so all $Q_i$ are zero.
Finally, we prove one last theorem:
Theorem 4: For all continuous $f : [a, b] \to \mathbb{C}$ where $a < b$, there is a sequence of polynomials $\{P_i\}_{i \in \mathbb{N}}$ with complex coefficients which uniformly converges to $f$ on $[a, b]$. If $f$ is real-valued, the $P_i$ may be taken to have real coefficients.
Proof: we can write $f(x) = g(x) + h(x) i$ where $g, h : [a, b] \to \mathbb{R}$ are both continuous. Thus, both $g$ and $h$ have the convergence property, so we can write $g$ as the uniform limit of $\{P_i\}_{i \in \mathbb{N}}$ and $h$ as the uniform limit of $\{Q_i\}_{i \in \mathbb{N}}$. Then $f$ is the uniform limit of $\{P_j + i Q_j\}_{j \in \mathbb{N}}$. Furthermore, if $f$ is real-valued, then $h$ is zero everywhere, so the $Q_j$ are zero everywhere, so $P_j + i Q_j$ is real everywhere and hence has real coefficients.
Best Answer
Note that $$ \int_{-1}^1 Q_n dx = c_n \underbrace{\int_{-1}^1 (1-x^2)^n dx}_{=:I_n}$$
If the integral term $I_n$ on the right was already $1$, then you can simply set $c_n =1$. But $I_n$ may not be $1$. By choosing $c_n$, we can change the value to anything we like, since $I_n$ is not zero. In particular we may make the choice
$$c_n = \frac1{I_n}=\frac1{\int_{-1}^1 (1-x^2)^n dx}.$$
The exact value of $I_n$ is not important. Only the much easier property that $I_n\neq0$. (Though you can with effort or a computer show that $I_n = \sqrt\pi \Gamma(n+1)/\Gamma(n+3/2)$.)