Proving $[0,1]^{\mathbb{N}}$ with the product topology is compact by showing it's complete and totally bounded The exercise:

Consider the Hilbert cube $C = [0,1]^{\mathbb{N}}$ with the metric that induces the product topology, i.e, $$d(x, y) = \displaystyle{\sup_{n \in \mathbb{N}}} \left\{\dfrac{\left|x_n - y_n \right|}{n} \right\}$$
a) Prove that in this topology, the open balls centered at $x$ with radius $r$, $B_d(x, r)$, are sets of the form $\displaystyle{\prod_{n \in \mathbb{N}} B(x_n, nr)}$ (where $B$ without the index stands for the balls of $[0,1]$ with the absolute value metric)
b) Prove that $C$ is complete
c) Check that $C$ is also totally bounded and conclude that it is compact

Now, here's my work (I've been cracking my head with this for a long time now, so that's why I'm posting it here: make sure I didn't make any mistakes and improve my work if possible):
a) is false and is not relevant to the other items. If we consider $B_d(\mathbf{0},1)$ and $\displaystyle{\prod_{n \in \mathbb{N}} B(0, n)}$, where $\mathbf{0} = (0, 0, \cdots)$, we see that if $y = (y_n)_{n \in \mathbb{N}}$ is defined by $y_n = \dfrac{n(2^n - 1)}{2^n}$, then $y \in \displaystyle{\prod_{n \in \mathbb{N}} B(0, n)}$ but $y \notin B_d(\mathbf{0},1)$.
b) Let $\{x_k \}_{k \in \mathbb{N}} \subset [0,1]^{\mathbb{N}}$ be a Cauchy sequence, with $x_k = (x_k^{(1)},x_k^{(2)}, \cdots) \in [0,1]^{\mathbb{N}}$. We want to show that $\{x_k \}_{k \in \mathbb{N}}$ converges, that is, to find a sequence $y = (y^{(1)}, y^{(2)}, \cdots) = (y^{(j)})_{j \in \mathbb{N}}$ such that $d(x_k, y) \to 0$. We can do this by showing that if $\{x_k \}_{k \in \mathbb{N}}$ is a Cauchy sequence, then $\{x_k^{(j)} \}_{k \in \mathbb{N}} = \{\pi_j(x_k) \}_{k \in \mathbb{N}} \subset [0,1]$ converges for each $j \in \mathbb{N}$. Then, using that convergence in the product topology is equivalent to convergence in each coordinate, we can conclude that $C$ is complete.
Indeed, if we are given arbitrary $\varepsilon > 0$ and $j_0 \in \mathbb{N}$, then, by hypothesis, there exists $K \in \mathbb{N}$ such that $k_1, k_2 \geq K$ implies the following:
$$d(x_{k_1}, x_{k_2}) = \displaystyle{\sup_{n \in \mathbb{N}}} \left\{\dfrac{\left|x_{k_1}^{(n)} - x_{k_2}^{(n)} \right|}{n}  \right\} < \dfrac{\varepsilon}{j_0}$$
Since it's also obvious that: $$\dfrac{\left|x_{k_1}^{(j_0)} - x_{k_2}^{(j_0)}\right|}{j_0} \leq \displaystyle{\sup_{n \in \mathbb{N}}} \left\{\dfrac{\left|x_{k_1}^{(n)} - x_{k_2}^{(n)} \right|}{n} \right\} < \dfrac{\varepsilon}{j_0} \implies \left|x_{k_1}^{(j_0)} - x_{k_2}^{(j_0)}\right| < \varepsilon$$
we conclude that $\{x_k^{(j)} \}_{k \in \mathbb{N}} $ is a Cauchy sequence in $[0,1]$ and therefore converges to some $y^{(j)}$. As desired, we have that $\pi_j(x_k) \to \pi_j(y)$ for each $j \in \mathbb{N}$, where $y = (y^{(1)}, y^{(2)}, \cdots) = (y^{(j)})_{j \in \mathbb{N}}$.
c) Let $\varepsilon > 0$ and take $n \in \mathbb{N}$ such that $\frac{1}{N} < \frac{\varepsilon}{2}$. For each $i > N$, take $p_i = 1$. Fix $i \in \{1, \cdots, N \}$. Since $[0,1]$ is totally bounded, there exist a finite number of points, say, $\{x_1, x_2, \cdots, x_M\} = A_0$ such that:
$$x \in [0, 1] \implies |x - x_j| < \frac{\varepsilon}{2} \text{ for some $j \in \{1, \cdots, M\}$}$$
Now, define $A = \{x = (x_n)_{n \in \mathbb{N}} \in [0,1]^{\mathbb{N}} \ \vert \ x_1, x_2, \cdots, x_N \in A_0 \text{ and } x_{N+1} = x_{N+2} = \cdots = 1  \}$. There are $M^N$ points in $A$, so $A$ is finite. Besides, if $x = (x_n)_{n \in \mathbb{N}} \in [0,1]^{\mathbb{N}}$, then for each $i \in \{1, \cdots, N\}$, there exist $x^{(i)}_j \in A_0$ (which means all the $x_j$ depend on the $i$) such that $|x_i - x^{(i)}_j | < \frac{\varepsilon}{2}$. Now, if we define $y_i = x^{(i)}_j $ for $i \in \{1, \cdots, N\}$ and $y_i = 1$ otherwise, then by construction $y = (y_n)_{n \in \mathbb{N}} \in A$, and for each $i \in \{1, \cdots, N\}$, we have that $\frac{|x_i - y_i|}{n} \leq |x_i - y_i| < \frac{\varepsilon}{2}$ and for each $i > N$, $\frac{|x_i - y_i|}{i} \leq \frac{1}{i} < \frac{1}{N} < \frac{\varepsilon}{2}$, so $\displaystyle{\sup_{n \in \mathbb{N}}} \left\{\dfrac{\left|x_n - y_n \right|}{n} \right\} \leq \frac{\varepsilon}{2} < \varepsilon$ and it follows that $[0,1]^{\mathbb{N}} = \displaystyle{\bigcup_{a \in A}B(a,\varepsilon)}$, as desired.
 A: (a) No, it's true. Remember that all of our coordinates have to be in $[0,1]$ - so as soon as $nr>1$, that radius becomes large enough that a ball centered anywhere in the interval contains the whole interval. Your choice of $y$ simply isn't in $C$.
I started this as a comment, but realized it's both long for that and important to the answer, so here it is:
Your issue is that the supremum of infinitely many numbers $\frac{|x_n-y_n|}{n}$ each strictly less than $r$ may be equal to $r$, putting that point outside the open ball. While that would be true for general numbers, our condition $0\le x_n,y_n\le 1$ saves us here - only finitely many terms of the sequence actually matter, and the supremum becomes a finite maximum.
Let $N$ be an integer greater than $\frac1r$. Then, for $n\ge N$, $\frac{|x_n-y_n|}{n}\le \frac1n\le \frac1N$, and the supremum of all the terms indexed by such $n$ is at most $\frac1N$. The overall supremum is thus the maximum of $N$ terms - $\frac{|x_n-y_n|}{n}$ for $n\in \{1,2,\dots,N-1\}$ and $\sup_{n\ge N}\frac{|x_n-y_n|}{n}$. Each of those $N$ terms is strictly less than $r$, so their maximum is as well.
Also, the relevance of this one? It's showing that the metric actually induces the product topology. That's not something we can take for granted; none of the usual norms induce the product topology, after all.
(b) Cauchy overall implies Cauchy in each coordinate, which implies convergence in each coordinate, which implies convergence overall. Looks good - and that last step to overall convergence is where we use the implication in (a) that this metric induces the product topology.
(c) Yes, you've got it.
