But how can we infer the existence of an unbounded continuous on the space knowing only about some sequence of points in this space
By using the sequence to construct an unbounded continuous function.
Since the sequence - call it $(x_n)_{n\in\mathbb{N}}$ - has no convergent subsequence, every point occurs only finitely many times in the sequence. Passing to a subsequence, we may assume all the $x_n$ are distinct.
For every $m\in\mathbb{N}$, the distance of $x_m$ to the rest of the sequence is positive,
$$\delta_m := \inf \left\{ d(x_m,x_k) : k \in \mathbb{N}\setminus \{m\}\right\} > 0,$$
for if $\delta_m = 0$, then $(x_n)$ would have a subsequence converging to $x_m$.
Now consider the functions
$$f_m(x) = \left(1 - \frac{3}{\delta_m}d(x_m,x)\right)^+,$$
where $u^+$ is the positive part of $u$, $u^+ = \max \{u,0\}$. These functions are continuous since the maximum of two continuous functions is continuous. The function
$$f(x) = \sum_{m=0}^\infty m\cdot f_m(x)$$
is, if well-defined, unbounded, since $f(x_m) \geqslant m$.
It remains to see that $f$ is well-defined and continuous. That follows if we can show that every point $x\in X$ has a neighbourhood on which at most one of the $f_m$ attains values $\neq 0$.
If $x = x_m$ for some $m\in\mathbb{N}$, then $f_k\lvert B_{\delta_m/3}(x_m) \equiv 0$ for all $k\neq m$: Suppose we had $f_k(y)\neq 0$ for some $y\in B_{\delta_m/3}(x_m)$ and some $k\neq m$. Then
$$\max \{\delta_k,\delta_m\} \leqslant d(x_m,x_k) \leqslant d(x_m,y) + d(y,x_k) \leqslant \frac{\delta_m}{3} + \frac{\delta_k}{3} \leqslant 2\frac{\max \{\delta_m,\delta_k\}}{3} < \max \{\delta_k,\delta_m\},$$
so that is impossible.
If $x \neq x_m$ for all $m$, the argument is similar. Let $\delta = \inf \{ d(x,x_n) : n\in\mathbb{N}\}$. Then $\delta > 0$ for otherwise the sequence would have a subsequence converging to $x$. Then on $B_{\delta/4}(x)$ at most one $f_m$ can attain a nonzero value. Suppose again it weren't so, and also $f_k$ attained a nonzero value there. Then we have $y,z \in B_{\delta/4}(x)$ with $f_m(y) \neq 0$ and $f_k(z) \neq 0$. That implies $\frac{3}{4}\delta < \frac{1}{3}\min \{ \delta_m,\delta_k\}$ since $d(y,x_m) \geqslant d(x,x_m) - d(x,y) \geqslant \frac{3}{4}\delta$ and similarly for $x_k$. But then we would have
$$0 < \max \{ \delta_m,\delta_k\} \leqslant d(x_m,x_k) \leqslant d(x_m,y) + d(y,z) + d(z,x_k) \leqslant \frac{\delta_m}{3} + \frac{1}{2}\delta + \frac{\delta_k}{3} < \max \{\delta_m,\delta_k\}.$$
Since every point has a neighbourhood on which at most one $f_k$ does not identically vanish, $f$ is well-defined and continuous.
The idea is naturally to construct a function which is continuous, but not bounded.
I will modify your construction (which seems to work, but is harder to analyze as the balls can intersect).
So we start with finding a sequence of non-intersecting balls. For $x_n$ and $r_n$ as in your answer, $B_n = B(x_n,r_n/2)$ are guaranteed to be disjoint. Now we construct a continuous function $f_n$ such that $f(x_n) = n$ and $f_n = 0$ outside $B_n$. This is easy, it is in fact constructed in your question, just replace $d(x,x_n)\le r_n$ with $d(x,x_n)\le r_n/2$. Finally, take $f(x) = \sum_{n\ge 1} f_n(x) \mathbf{1}_{B_n} (x)$. Then $f$ is continuous but unbounded.
Best Answer
Let $(s_n)_{n\in\Bbb N}$ be a sequence of elements of $S$ which converges to some element $x\in X$. Then each $s_n$ is some $x_k$. Since $x$ is not the limit of the sequence $(x_n)_{n\in\Bbb N}$, then there is some open ball $B_r(x)$ with no $x_n$ in it, with the possible exception of $x$ itself. But, since $\lim_{n\to\infty}s_n=x$, this would mean that $s_n=x$ if $n\gg1$.
This proves that the limit of every convergent sequence of elements of $S$ is an element of $S$. Therefore, $S$ is a closed set.
And it is a discrete set because, if $x_N\in S$ was not an isolated point of $S$, then there would be a sequence $(s_n)_{n\in\Bbb N}$ that would not be eventually constant and that would converge to $x_N$. But it was already proved that such a sequence does not exist.