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.
If a metric space $X$ is non-compact then $X$ has a countably infinite closed discrete subspace Y. Because if $(p_n)_{n\in \mathbb N}$ is a Cauchy sequence in $X$ with no convergent subsequence, let $Y=\{p_n:n\in \mathbb N\}.$
We show that if $(X,d)$ is connected and non-compact then there is a continuous $g:X\to \mathbb R$ that is not uniformly continuous.
Let $Y=\{y_n:n\in \mathbb N\}$ be a closed discrete subspace of $X$ such that $m\ne n\implies y_m\ne y_n.$ For each $n$ let $r_n=\inf \{d(y_n,y_m):m\ne n\}.$ We have $r_n>0.$ Let $s_n=\min (1/n, r_n/6).$
Let $f_n:X\to [0,1]$ be continuous with $f(y_n)=1,$ and $f(x)=0$ for all $x\in X$ \ $B_d(y_n,s_n).$
The triangle inequality and the def'n of $r_n$ imply that for any $x\in X$ there exists $t_x>0$ such that there is at most one $n$ such that $B_d(x,t_x)\cap B_d(y_n,s_n)\ne \phi.$ So $f_n(x)\ne 0$ for at most one $n.$ So $g(x)=\sum_{n=1}^{\infty}f_n(x)$ is defined.
For brevity (of subscripts) let $U_x=B_d(x,t_x).$ Now $g(x)$ is locally continuous: For each $x\in X$ the function $g|_{U_x}$ is either constantly $0$ or is $f_n|_{U_x}$ for some (unique) $n.$ Local continuity everywhere is equivalent to continuity. So $g$ is continuous.
Now $X$ is connected so each non-empty open set $B_d(y_n,s_n)$ is not closed. (Its complement is not empty as it contains $y_{n+1}).$ So let $z_n$ belong to the boundary of $B_d(y_n,s_n).$ We have $g(y_n)=f_n(y_n)=1$ and $g(z_n)=f_n(z_n)=0$ but $d(y_n,z_n)=s_n\leq 1/n.$ So $g$ cannot be uniformly continuous. QED.
APPENDIX: To prove that for $x\in X$ there exists $t_x>0$ such that there is at most one $n$ such that $B_d(x,t_x)\cap B_d(y_n,s_n)\ne \phi:$
Note that for $n\ne n'$ we have $r_n\leq d(y_n,y'_n)\geq r_{n'}$ so $d(y_n,y_{n'})\geq \max(r_n,r'_n).$
(1). If $n\ne n'$ then $B_d(y_n,s_n)$ and $B_d(y_{n'},s_{n'})$ are disjoint, because if $z$ is in both of them then $$0<\max (r_n,r_{n'})\leq d(y_n,y_{n'})\leq d(y_n,z)+d(z,y_{n'})<s_n+s_{n'}\leq r_n/6+r_{n'}/6\leq$$ $$\leq \max (r_n,r_{n'})/3$$ a contradiction.
So if $x\in B_d(y_n,s_n)$ for some $n,$ we can take $t_x>0$ where $t_x$ is small enough that $B_d(x,t_x)\subset B_d(y_n,s_n).$
(2). If $x\not \in B_d(y_n,r_n)$ for any $n,$ let $t_x=(1/3)\inf \{d(x,y):y\in Y\}.$ We have $t_x>0$. Suppose by contradiction that $n\ne n',$ with $a\in B_d(y_n,s_n)\cap B_d(x,t_x)$ and $b\in B_d(y_{n'},s_{n'})\cap B_d(x,t_x).$ Then $$d(y_n,y_n')\leq d(y_n,a)+d(a,x)+d(x,b)+d(b,y_{n'})<s_n+2t_x+s_{n'}\leq$$ $$\leq r_n/6+2t_x+r_{n'}/6\leq 2t+d(y_n,y_{n'})/3$$ which implies $t_x\geq d(y_n,y_{n'})/3\geq r_n/3\geq 2s_n.$ This implies $$d(x,y_n)\leq d(x,a)+d(a,y_n)\leq t_x+s_n\leq t_x+(t_x/2)<2t_x.$$ So $y_n\in B_d(x,2t_x),$ contradicting the def'n of $t_x.$
Best Answer
Let
Then $f$ is not continuous, otherwise, $f$ restricted to the diagonal $\Delta$ of $X\times Y$ will be continuous. But $f(x,x)=\frac{1}{2}$ for $(x,x)\neq (0,0)$; Thus $(f|_{\Delta})^{-1}(\frac{1}{2})=\Delta \setminus \{(0,0)\}$, which is not closed in $\Delta$, though $\{\frac{1}{2}\}$ is closed in $Z$.
Lemma 1:
proof: $f(\Bbb Q)$ is countable, hence closed in $\Bbb R_{\text{cocountable}}$. $f^{-1}(f(\Bbb Q))$ is a closed set containing $\Bbb Q$, hence equal to $\Bbb R$. $f^{-1}(f(\Bbb Q))=\Bbb R$ implies $f(\Bbb R)=f(f^{-1}(f(\Bbb Q)))\subseteq f(\Bbb Q) $, hence $f(\Bbb R)$ is countable. A countable, connected subset of $\Bbb R_{\text{cocountable}}$ must be a one-point set. Hence $f$ is constant. $\square$
Claim 1:
proof: $\pi_2\circ g$ is a continuous function from $\Bbb R$ to $\Bbb R_{\text{cocountable}}$. ($\pi_2$ is the projection onto the second coordinate). By lemma 1, $\pi_2\circ g$ is constant. Hence the result follows. $\square$
Hence for every continuous function $g: \Bbb R \to \Bbb R\times \Bbb R_{\text{cocountable}}$, the precomposition $f\circ g:\Bbb R\to \Bbb R_{\text{cofinite}}$ is equal to $f(x,y_0)=\frac{x^2y_0^2}{x^2+y_0^2}$ ($y_0$ fixed). Replacing $\Bbb R_{\text{cofinite}}$ by the finer topology $\Bbb R$, the function is easily seen to be continuous by standard calculus argument. Hence $f\circ g:\Bbb R\to \Bbb R_{\text{cofinite}}$ is continuous.
Lemma 2:
proof: (continuity on co-countable topology drhab's answer) Let $h$ be such a function with e.g. $0,1\in h\left(\mathbb{R}\right)$. Suppose $f$ is continuous and let $D_0,D_1$ be disjoint open sets containing $0$ and $1$ respectively. Then $h^{-1}\left(D_0 \right)$ and $h^{-1}\left(D_1\right)$ must be disjoint sets both having a countable complement. Then $h^{-1}\left(D_0\right)$ as a subset of the complement of $h^{-1}\left(D_1\right)$ is countable and consequently $\mathbb{R}=h^{-1}\left(D_0 \right)\cup h^{-1}\left(D_0\right)^{c}$ is countable. Contradiction. $\square$
Claim 2:
proof: $\pi_1\circ h$ is a continuous function from $\Bbb R_{\text{cocuntable}}$ to $\Bbb R$. ($\pi_1$ is the projection onto the first coordinate). By lemma 2, $\pi_1\circ h$ is constant. Hence the result follows. $\square$
Hence for every continuous function $h: \Bbb R_{\text{cocountable}} \to \Bbb R\times \Bbb R_{\text{cocountable}}$, the precomposition $f\circ h:\Bbb R_{\text{cocountable}}\to \Bbb R_{\text{cofinite}}$ is equal to $f(x_0,y)=\frac{x_0^2y^2}{x_0^2+y^2}$ ($x_0$ fixed). $(f\circ h)^{-1}(A)$ is closed (i.e. countable) for every closed (i.e. finite) set $A$. In fact, the preimage of each element can have cardinality at most $2$ (by looking at the $\text{graph}^1$). Hence $f\circ h:\Bbb R_{\text{cocountable}}\to \Bbb R_{\text{cofinite}}$ is continuous.
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