Prob. 6, Chap. 1, in Rudin's PMA Here's problem $6$ in Chapter $1$ in the book Principles of Mathematical Analysis by Walter Rudin, $3$rd edition: 
Fix a real number $b$, such that $b > 1$. 
$(a)$ If $m, n, p, q$ are integers, $n > 0$, $q > 0$, and $r = m/n = p/q$, then (I've managed to show that) 
$$\left( b^m \right)^{1/n} = \left( b^p \right)^{1/q}.$$
Hence it makes sense to define
$$b^r \colon= \left(b^m\right)^{1/n}.$$
$(b)$ (I've also managed to show that) for any rational numbers $r, s$, we have 
$$b^{r+s} = b^r b^s.$$
$(c)$ For any real number $x$, define the set $B(x)$ as follows:
$$B(x) \colon= \left\{ \ b^t \ \colon \ t \ \mbox{ is rational}, \ t \leq x \ \right\}.$$
We can then prove that 
$$b^r = \sup B(r)$$
when $r$ is a rational number. 
Hence it makes sense to define 
$$b^x \colon= \sup B(x) = \sup \left\{ \ b^t \ \colon \ t \ \mbox{ is rational}, \ t \leq x \ \right\}$$
for every real number $x$. 
$(d)$ How to prove (USING THE MACHINERY DEVELOPED HERE) that $$b^{x+y} = b^x b^y$$
for all real numbers $x$ and $y$?
I've already posted this question. Here's the link. However, none of the answers there seems to work for me.

My Work 
Let  $X, Y$ be any two non-empty subsets of $\mathbb{R}$. Then we have the following facts: 
$(1)$ If $X \subset Y$ and if $Y$ is bounded above in $\mathbb{R}$, then so is $X$ and we have the inequality
$$\sup X \leq \sup Y.$$
$(2)$ If $X, Y$ are non-empty subsets of the set of non-negative real numbers and if both $X$ and $Y$ are bounded above, then so is the set
$$\{ \ xy \ \colon \ x \in X, \ y \in Y \ \};$$
moreover, we have the identity
$$\sup \{ \ xy \ \colon \ x \in X, \ y \in Y \ \} = \sup X \cdot \sup Y.$$
Thus,
$$\begin{align}
b^x b^y &= \sup B(x) \sup B(y) \\
&= \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, r \leq x \ \} \sup \{ \ b^s \ \colon \ s \in \mathbb{Q}, s \leq y \ \} \\
&= \sup \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \  r \leq x, \ s \leq y \ \}.
\end{align}$$
But if $r \in \mathbb{Q}$, $s \in \mathbb{Q}$, $r \leq x$, and $s \leq y$, then the sum $r+s \in \mathbb{Q}$ and $r + s \leq x + y$ also. So 
$$ \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \  r \leq x, \ s \leq y \ \} \subseteq \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x+y \ \}.$$
Therefore,
$$\begin{align}
b^x  b^y &= \sup \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \  r \leq x, \ s \leq y \ \} \\
&\leq \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x+y \ \} \\
&= \sup B(x+y) = b^{x+y}.
\end{align}$$
Now we have to show that
$$b^x  b^y \geq b^{x+y}.$$
Case I. When $x+y$ is irrational: 
If $t$ is a rational number such that $t \leq x+y$, then since $x+y$ is irrational, we can also conclude that $t < x+y$. 
So we can find a rational number $r$ such that
$$ t-y < r < x.$$
Then 
$$t-r < y < x+y -r.$$
Let's take $s \colon = t-r$. Then $s \in \mathbb{Q}$. And $s < y$. 
Thus, we have written $t$ as $t = r + s$, where $r, s \in \mathbb{Q}$, with $r < x$, $s < y$. 
So, when $x + y \in \mathbb{R} - \mathbb{Q}$, then we have 
$$\begin{align}
b^{x+y} &= \sup B(x+y) = \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x+y \ \} \\
&= \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t < x+y  \} \\
&= \sup \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r< x, \ s < y  \}  \\
&= \sup \{ \ b^r b^s  \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r< x, \ s < y  \}  \\
&= \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, \ r < x \ \} \sup  \{ \ b^s \ \colon \ s \in \mathbb{Q}, \ s < y  \} \\ 
&\leq \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, \ r \leq x \ \} \sup  \{ \ b^s \ \colon \ s \in \mathbb{Q}, \ s \leq y  \} \\ 
&= \sup B(x) \sup B(y) \\ 
&= b^x b^y.
\end{align}$$
Case II. What if $x + y \in \mathbb{Q}$? 
In this case, we can conclude that both $x$ and $y$ are irrational. 
How do we proceed with showing the remaining reverse inequality in this particular case?
 A: This is another instance of proving stuff with one hand tied behind your back, limiting knowledge of the reals to what is in Rudin ch. 1.  For this reason I write out a complete proof, even though most of it is given by the OP.  We assume $b>1$ real.
Lemma 0.  If $p>0$ is rational, $b^p>1$.
Lemma 1.  If $s<x+y$, $s$ rational, then $s=p+q$ with $p$ and $q$ rational, $p<x$, $q<y$.   
Proof.  Pick $s'$ rational with $s<s'<x+y$.  Pick $p$ rational with $x-(s'-s)<p<x$.  Let $q=s-p$. We then have $q=s-p <s-x+(s'-s)<s-x+x+y-s=y$ as desired.
Lemma 2 (Rudin ch. 1 ex. 1.7(c)).  If $t>1$ and $n>(b-1)/(t-1)$ then $b^{1/n}<t$.
Lemma 3. For any $x$ we have $b^x=\sup \{b^r:r<x, r\in\mathbb{Q}\}$. 
Proof.  If $x$ is irrational this follows from the definition and the fact that $r\ne x$ for rational $r$, so the set of rational $r\le x$ is really the set of rational $r<x$.  If $x$ is rational, by Lemma 0 $b^x/b^r=b^{x-r}>1$ so $b^x$ is an upper bound of $\{b^r: r<x, r\in\mathbb{Q}\}$.  Suppose there were a smaller upper bound $u$.  Choose $n$ so that $b^{1/n}<b^x/u$, using Lemma 2.  We then have $b^{x-1/n}>u$, with $x-1/n$ rational and $<x$, contradiction. 
Lemma 4.  If $X$ and $Y$ are sets of reals, $\sup X\cdot\sup Y = \sup \{xy:x\in X, y\in Y\}$.  (Stated to be proven by the OP.)
Now we're ready for the main result, $b^{x+y}=b^x b^y$. 
$$b^{x+y}=\sup\{b^r: r\in\mathbb{Q}, r<x+y\}=\sup\{b^{p+q}:p,q\in\mathbb{Q}, p<x, q<y\} =\\ \sup\{b^p:p\in\mathbb{Q}, p<x\}\cdot\sup\{b^q:q\in\mathbb{Q}, q<y\}=b^xb^y$$
The first and last equalities follow from Lemma 3.  The second follows from Lemma 1.  The third follows from Lemma 4.  Fingers crossed I haven't made a mistake.
