This is not meant to replace azarel’s answer, but rather to show in more detail just how unbounded real-valued functions arise in non-compact metric spaces.
Recall that if a metric space $\langle X,d\rangle$ is not compact, then either it’s not complete, or it’s not totally bounded.
If it’s not complete, it has a Cauchy sequence $\langle x_n:n\in\omega\rangle$ that does not converge; define $f:X\to\mathbb{R}:x\mapsto\lim_{n\to\infty}d(x,x_n)\;.$
Intuitively, $f(x)$ is the distance from $x$ to the non-existent point to which the Cauchy sequence wants to converge, so you’d expect $f$ to exist and be continuous, and indeed this isn’t hard to prove. Moreover, $f(x)$ is never $0$, because the Cauchy sequence has no limit in $X$. Thus, the function $g:X\to\mathbb{R}:x\mapsto\frac1{f(x)}$ is continuous, and it’s also clearly unbounded, since $\lim_{n\to\infty}g(x_n)=\infty\;.$ In effect we’ve pushed that non-existent limit of the Cauchy sequence out to be a point at infinity.
If $X$ is not totally bounded, there is an $r>0$ such that no finite collection of open $r$-balls covers $X$. Thus, we may recursively construct a sequence $\langle x_n:n\in\omega\rangle$ such that for each $n\in\omega$, $x_n\in X\setminus\bigcup_{k where $B(x_k,r)$ is the open ball of radius $r$ centred at $x_k$. Note that $d(x_m,x_n)\ge r$ whenever $m, so $\{B(x_n,r/3):n\in\omega\}$ is an infinite family of pairwise disjoint open balls of radius $r/3$. Now let
$g:X\to\mathbb{R}:x\mapsto\begin{cases} 0,&\text{if }x\in X\setminus\bigcup_{n\in\omega}B\left(x_n,\frac{r}3\right)\\\\ \frac{3n}r\left(\frac{r}3-d(x,x_n)\right),&\text{if }x\in B\left(x_n,\frac{r}3\right)\;. \end{cases}$
Clearly $g(x_n)=n$ for each $n\in\omega$, so $g$ is unbounded, and once again it isn’t hard to show that $g$ is continuous.
This time the intuitive idea is to replace the continuous function $B(x,s)\to\mathbb{R}:y\mapsto d(x,y)$ that measures distance from the centre of a ball with the function $B(x,s)\to\mathbb{R}:y\mapsto s-d(x,y)\tag{1}$ that in a sense measures distance from the boundary of the ball, and then ‘blow up’ the function in $(1)$, making the ball’s radius look bigger than $s$. Since we have infinitely many balls to play with, and they don’t interfere with one another, we can get arbitrarily large ‘blow-up’.