Let $B = \{f \in \mathcal{C}([0,1], \mathbb{R}) : f$ is nondecreasing$\}$. Let $d_2$ be the $2$-metric on $\mathcal{C}([0,1], \mathbb{R}),$ defined so that $d_2(f,g) = (\int_0^1 (f(x)-g(x))^2dx)^{1/2}$.
Is $(B, d_2)$ separable? And if so, how can I show this? Or how can I provide a counterexample? I don't want to use the Stone-Weierstrass theorem and I believe it shouldn't be necessary given the definition of $B$. I think that polynomials will be useful if it's separable.
Also, is $(B, d_2)$ complete? If I could show that $(\mathcal{C}([0,1]), d_2)$ is complete, then it would suffice to show that $B$ is closed. So let $(f_n)\subseteq B, f_n\to f.$ Then $(\int_0^1 (f_n(x)-f(x))^2dx)^{1/2}\to 0\Rightarrow \int_0^1 (f_n(x)-f(x))^2dx\to 0,$ and I think one can show from this that $f_n\to f$ pointwise, but I'm not really sure how to do this exactly. However, if I can do this, the rest seems relatively straightforward. Let $x\leq y \in [0,1].$ Observe that $f(x)\leq f(y)$ as $f_n(x)\leq f_n(y)$ for all $n$ and $f$ is the pointwise limit of $(f_n).$
Clarification: To justify that the set is separable, I would like an answer that goes into detail about how uniform continuity can be leveraged to uniformly approximate nondecreasing functions using piecewise functions (i.e. use the definition of an integral to explicitly construct a piecewise linear continuous nondecreasing function that gets arbitrarily close to an arbitrary nondecreasing function in $B$).
Not a solution, but a sketch to the first question: Consider the sets of finite sequences of points $$S_n:=\left\{\left((0,y_0),\ldots, (\frac{k}{n},y_k),\ldots, (1, y_n)\right): y_k\in\mathbb{Q}, 0\le y_k\le y_{k+1}\le1 \right\}$$ $S_n$ is obviousy countable, and each sequence in $S_n$ gives rise to a piecewise linear function by linear interpolation from $(0,y_0)$ through the elements of such sequence to $(1,y_n)$. Denote the set of these functions by $PL_{n}$ - clearly this set is also countable
An element of $PL_n$ is, by construction, continuous and nondecreasing, so it is an element of your set $B$. It is not difficult to see that every $f\in B$ can be uniformly approximated by such piecewise linear functions. (For the proof of ths claim this you need to know that a continuous function on a closed bounded interval is uniformly continuous).
If you now know that each function $f$ in $B$ can be uniformly approximated with functions from the countable set $PL_n$, all which remains to show is that uniform convergence implies $d_2$ convergence. But
$$d_2(f,g)^2 = \int_0^1(f-g)^2(t)dt \le ||f-g||_\infty^2$$ so that should be obvious.
Regarding your second point note that the counterexample I outlined will also show that $B$ is not complete wrt to $d_2$, since the sequence proposed can be chosen as a sequence of nondecreasing functions, i.e. functions taken from $B$.
Here is a proof which shows that each $f\in B$ can be approximated by a function from some $PL_n$. Assume $f\in B$ and $\varepsilon > 0 $ is given. Since $f $ is continuous and $[0,1]$ compact, $f$ is uniformly continuous, i.e. there is a number $N\in \mathbb{N}$ such that for every $n\ge N$
\begin{equation} |x-y| < \frac{1}{n} \quad \Longrightarrow |f(x) - f(y) |< \varepsilon \end{equation}
Choose any $n\ge N$ and for each number $x_k:= \frac{k}{n}\, (1\le k \le n)$ choose a rational number $q_k$ such that $$|f(x_k) - q_k| < \varepsilon$$ This is possible since $\mathbb{Q}$ is dense in the real numbers.
Since $f$ is nondecreasing this is true also for the sequence $f(x_k)$ and it is not difficult to see that then the sequence of the $q_k$ can also be chosen to be nondecreasing, i.e. $q_k\le q_{k+1}$ for every $0\le k <n$ .
As discussed earlier, the set of $q_k$ gives rise to a function $h \in PL_n$ by $h(x_k) = q_k $ and extending $h$ linearly on each subinterval $[x_k , x_{k+1}]$. Now chose $x\in [0,1]$ and $k$ such that $x_k\le x \le x_{k+1}$. Then \begin{eqnarray} |f(x) - h(x) | & = & |f(x) - f(x_k) + f(x_k) - h(x_k) + h(x_k) - h(x)| \\ & \le& |f(x) - f(x_k)| + |f(x_k) - h(x_k)| + |h(x_k) - h(x)| \\ & \le & \varepsilon + \varepsilon + 3\varepsilon = 5\varepsilon \end{eqnarray} Here, the first term from the line in the middle, was estimated using the uniform continuity of $f$ and the second one by construction of the function $h$, sinc $h(x_k) = q_k$. The last term was estimated with the help of
\begin{eqnarray} |h(x_{k+1})-h(x_k)| & = & |h(x_{k+1}) -f(x_{k+1}) +f(x_{k+1}) - f(x_k) + f(x_k)-h(x_k)| \\ & \le& |h(x_{k+1}) -f(x_{k+1}) | + | f(x_{k+1}) - f(x_k) | + | f(x_k)-h(x_k) | \\ &<& 3 \varepsilon \end{eqnarray} and the fact that $$|h(x)- h(x_k)| \le |h(x_{k+1})-h(x_k)| $$ which follows from the linearity of $h$ on the interval $[x_k,x_{k+1} ]$
So we see that, for every $\varepsilon >0$ there is $h\in PL_n$ (with $n\ge N = N(\varepsilon)$) which is $5\varepsilon $ close to $f$.