Part of the argument is simply wrong: the base constructed is not countable, because it uses all $\epsilon>0$ instead of a countable set of $\epsilon>0$ containing arbitrarily small members, like $\{2^{-n}:n\in\Bbb N\}$ or $\left\{\frac1n:n\in\Bbb Z^+\right\}$. The intended argument could be made much more clearly. Let me simply do a more intelligible version, rather than comment specifically on this one.
For each $n\in\Bbb N$, $\{B(x,2^{-n}:x\in X\}$ is an open cover of the Lindelöf space $X$, so it has a countable subcover $\mathscr{B}_n$. Let $\mathscr{B}=\bigcup_{n\in\Bbb N}\mathscr{B}_n$; $\mathscr{B}$ is a countable family of open subsets of $X$, and I claim that it’s a base for the topology of $X$.
To see this, let $U$ be any non-empty open set in $X$, and fix $x\in U$; we must show that there is some $B\in\mathscr{B}$ such that $x\in B\subseteq U$. Since $x\in U$, and $U$ is open, we know that there is an $\epsilon>0$ such that $B(x,\epsilon)\subseteq U$. Choose $n\in\Bbb N$ large enough so that $2^{-n}<\frac{\epsilon}2$. Now $\mathscr{B}_n$ covers $X$, so there is a $B(y,2^{-n})\in\mathscr{B}_n$ such that $x\in B(y,2^{-n})$; I claim that $B(y,2^{-n})\subseteq U$.
Suppose that $z\in B(y,2^{-n})$; then $d(z,y)<2^{-n}<\frac{\epsilon}2$. We also know that $d(x,y)<2^{-n}<\frac{\epsilon}2$, so by the triangle inequality we have $d(x,z)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$, $z\in B(x,\epsilon)\subseteq U$, and hence $B(y,2^{-n})\subseteq U$, as claimed. It follows that $\mathscr{B}$ is indeed a base for the topology on $X$.
Added: Getting back to your specific questions about the original form of the argument, I think that what you’ve not realized is that for each $\epsilon>0$ there is only one family $\{B(x_{\epsilon,n},\epsilon:n\in\Bbb N\}$ of $\epsilon$ balls being considered; there is not a separate one for each $B(y,d)$. (The double subscript on $x_{\epsilon,n}$ is necessary, because for each $\epsilon>0$ there is potentially a different $x_n$.) We use the Lindelöf property to find these countable covers of $X$ at the beginning; we don’t find new ones for each $B(y,d)$. Part of the problem, I think, is that the argument given does not adequately explain why we can be certain that there is a $B(x_{\epsilon,n},\epsilon)$ such that $d(x_n,y)+\epsilon<d$ and $\epsilon\ge d(x_{\epsilon,n},y)$.
The Sorgenfrey line, also called the lower-limit topology on the real line. It has a basis of intervals $[a,b)$ (or some authors prefer $(a,b]$, upper-limit topology).
It is hereditarily Lindelof (and hereditarily separable, i.e. every subspace has a countable dense subset), but it is not second countable.
Real line with discrete topology is not Lindelof.
$(0,1)$ with the usual topology is Lindelof, it is homeomorphic to the real line.
The real line is $\sigma$-compact, that is union of countably many compact subspaces. The real line is the union $\cup_n[-n,n]$, clearly $\sigma$-compact.
The interval $(0,1)=\cup_n [\frac1n,1-\frac1n]$ is $\sigma$-compact.
Every $\sigma$-compact space is Lindelof (easy to prove).
In particular, clearly every countable space is Lindelof. But not every countable space is second countable (even with nice separation axioms). For example, the countable sequential fan is not second countable. It is obtained by taking a disjoint family of countably many convergent sequences, and
"gluing" their limit points into one limit point, via a quotient map. The result is not first countable at that point. (I couldn't find a suitable online reference to the countable sequential fan, but it has similar properties to the quotient space $\Bbb R/\Bbb N$, which is also not first countable, and likely discussed in most topology books.)
There is an online searchable database (called $\pi$-base), you can make a query asking for Lindelof, not second countable spaces. For many more examples see
https://topology.jdabbs.com/spaces?q=lindelof%20%2B%20~Second%20Countable
Every second countable space is Lindelof (you may need to assume some separation axioms, often included in the definitions).
One of the examples at $\pi$-base is the one-point Lidelofication of uncountable discrete space. Take any uncountable set, and a point $p$, and isolate all point but $p$. The neighborhoods of $p$ are co-countable (that is, they have a countable complement). The definition easily implies that the result is a Lindelof space. But it is not first-countable at $p$, and hence not second countable. Often (in the case when "uncountable" is taken to be the first uncountable cardinal) this space is described as the set of all countable ordinals $\omega_1=\{\alpha:\alpha<\omega_1\}$ together with the first uncountable ordinal $\omega_1$, so $X=[0,\omega_1]=\{\alpha:\alpha\le\omega_1\}$, with all countable ordinals $\alpha<\omega_1$ isolated, and with basic neighborhoods of $\omega_1$
of the form $(\alpha,\omega_1]$, with $\alpha<\omega_1$. This is, in addition, an example of a Linearly Ordered Topological Space (LOTS), with only one non-isolated point. (It is a LOTS under a somewhat different order, one may insert a decreasing sequence in front of every limit ordinal.)
The last (and some of the previous examples) are not second countable, because they are not even first countable (and, for the one-point Lindelofication the point $p$ is not even a $G_\delta$ point, that is, it is not the intersection of any countable family of open sets, and the space is not $\sigma$-compact). On the other hand the Sorgenfrey line is perfectly normal: Every closed set (and in particular every point) is the intersection of a countable family of open sets (the proof that the Sorgenfrey line is hereditarily Lindelof uses this). But it is not second countable, since for every basis, and for every $x$ there must be a basic element $B_x$ with $x\in B_x\subseteq[x,\infty)$ and clearly if $x\neq y$ then $B_x\neq B_y$.
An example of a space that is separable but not second countable and not Lindelof is the Moore, or Niemytzki plane (also called tangent-disk space, usually available in topology texts).
There are also compact spaces (which of course is stronger than Lindelof) that are not second countable. One such example closely related to the Sorgenfrey line is the Alexandrov double arrow space, also called split interval.
Another is the Alexandroff double circle. Note that every second countable space in hereditarily Lindelof (since every subspace is second countable, and hence Lindelof). The Alexandroff double circle is compact (and hence Lindelof), but has an uncountable discrete subspace, which of course is not Lindelof. Hence the Alexandroff double circle is not second countable.
Best Answer
Let $\mathcal{U}$ be an open cover of $(X,d)$. Let $\{d_n: n \in \mathbb{N}\}$ be a dense subset of $X$. For every $n\in \mathbb{N}, q \in \mathbb{Q}$, if there exists some member $U$ of $\mathcal{U}$ that contains $B(d_n, q)$, pick some $U(n,q) \in \mathcal{U}$ that does (there could be plenty of such $U$ so we use AC to pick definite ones). Otherwise $U(n,q) = U_0$ for some fixed $U_0 \in \mathcal{U}$, for definiteness.
Claim: $\{ U(n,q): n \in \mathbb{N}, q \in \mathbb{Q}\}$ is a countable subcover of $\mathcal{U}$.
To see this, let $x \in X$ and find some $U_x \in \mathcal{U}$ that contains $x$ (as we have a cover). Then for some $r>0$ we have $B(x,r) \subset U_x$, and we can also find some $d_{n(x)}$ from the dense subset inside $B(x,\frac{r}{2})$, and next we find a $q(x) \in \mathbb{Q}$ such that $d(x, d_{n(x)}) < q(x) < \frac{r}{2}$.
Note that $x \in B(d_{n(x)}, q(x))$ and $B(d_{n(x)}, q(x)) \subset B(x,r) (\subset U_x)$ by the triangle inequality.
Then $U_x$ witnesses that some member of $\mathcal{U}$ contains $B(d_{n(x)},q)$ and so we know that $x \in B(d_{n(x)}, q(x)) \subset U(n(x), q(x))$ and so $x$ is indeed covered by the subcover, as required.
This proof is quite direct, but it is essentially the same argument one needs to see that all sets $B(d_n, q)$ for $n \in \mathbb{N}, q \in \mathbb{Q}$ form a (countable) base for $X$. So it's not really a simplification per se, but just to show a direct proof is possible.
BTW, it's not too hard to go to hereditarily Lindelöf as well. Essentially the same proof holds: $\mathcal{U}$ is then not necessarily a cover of $X$ and the $x$ is then chosen to be in $\cup \mathcal{U}$ instead of $X$. We use the formulation of hereditarily Lindelöf as : every family of open sets has a countable subfamily with the same union. The proof goes through the same way, really.