Let $S$ be a countable set of real numbers that is bounded but has neither a maximum nor a minimum. Next we create the product topology $[0, 1]^S$ (using the usual topology on $[0, 1]$). This should be compact since $[0, 1]$ is. Now we create a subspace $P$ of $[0, 1]^S$. A member of $[0, 1]^S$ is in $P$ iff the sum of its components is $1$. (Essentially, $P$ is the set of probability distributions with domain $S$.) $P$ should be closed (since it is essentially a surface), and therefore compact.
Now we create a continuous surjection $f : P \to (a, b)$, where $a$ is the infinum and $b$ is the supremum of $S$. We define $f$ by $$f(p) = \sum_{s \in S} p_ss$$ (i.e. $f(p)$ is the "expected value" of $p$).
This always falls in the range $(a, b)$ since for any $p \in P$, there must be some $s \in (a, b)$ such that $p_s > 0$. Then $b < p_ss + (1-p_s)b \le f(p) \le p_ss + (1 - p_s)a < a$.
Additionally, $f$ is surjective, since for any $r \in (a, b)$, we just take some $s_1$ and $s_2$ such that $a < s_1 < r < s_2 < b$ and define $p_{s_1}$ as $\frac{s_2 - r}{s_2 - s_1}$, $p_{s_2}$ as $\frac{r - s_1}{s_2-s_1}$, and $p_{s'}$ as $0$ for all other $s' \in S$. Then $p \in P$ and $f(p) = r$.
Since $f$ is surjective and $P$ is compact, $(a,b)$ is compact. But $(a,b)$ is not compact, which is a contradiction!
What did I do wrong?
I suspect the error is either that $P$ is not closed in $[0, 1]^S$ or $f$ is not continuous. I'm pretty sure all the other steps are correct.
The problem arises from the fact that $P$ is not closed in $[0,1]^S$. Let us write $S=\{s_n\}_{n=1}^\infty$ and let $p_n(s_m)=1$ for $n=m$ and $0$ otherwise. Then, $p_n$ converges to $p\equiv 0$ in product topology but $p\notin P$. Intuitively, $P$ is the space of all probability measures on $S$. To make $P$ compact, we should give $P$ another topology such as weak topology (topology induced from weak convergence of measure).