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.
Best Answer
Let $K$ be a set which is such that every continuous function on it is bounded.
Clearly $K$ itself is bounded, for the function «distance to the origin» is bounded by hypothesis.
Suppose $K$ is not closed, so that there is a point $x$ which is in the closure of $K$ but not in $K$. Consider the function $$f:y\in K\mapsto \frac{1}{d(x,y)}\in\mathbb R,$$ which is clearly well defined and continuous. The choice of $x$ implies more or less immediately that $f$ is not bounded on $K$, so something's amiss...