For a group $G$, let $F(G)$ denote the collection of all subgroups of $G$. Which one of the following situations can occur?
a) $G$ is finite but $F(G)$ is infinite
b) $G$ is infinite but $F(G)$ is finite
c) $G$ is countable but $F(G)$ is uncountable
d) $G$ is uncountable but $F(G)$ is countable
a) $G$ is finite means $\vert G \vert =n$, a finite number. so it has finite number of subgroups. so a) is false
c) can occur as this post shows.
My try for d):
take $G=C[0,1]$, the ring of all real valued continuous functions on $[0,1]$. It is a ring and in particular it is an uncountable additive group. It has uncountable number of maximal ideals, namely of the form $$H_\gamma=\{f \in C[0,1] : f(\gamma)=0\}$$ where $\gamma$ is any number in $[0,1]$.
These Uncountable $H_\gamma$'s are in particular additive subgroups of $C[0,1]$. So d) is false
My question is :
1) How to disprove b) ?
that is., How to prove an infinite group has infinite number of subgroups ?
2) Is my counterexample valid?
Your example shows that (d) doesn't always hold, not that it is false.
To show (b) is false, let $G$ be a countable group. If $G$ contains an element $g$ of infinite order, then $\langle g^n\rangle$, $n\geq 1$ gives an infinite number of subgroups.
Therefore, assume that every element of $G$ has finite order. Then, the set $F_{\mathrm{fin}}(G)$ of finite subgroups of $G$ is already infinite. Indeed, $\{\langle g\rangle\mid g\in G\}$ must be an infinite set. If it were finite, then $G=\bigcup_{g\in G}\langle g\rangle$ would be finite being the finite union of finite sets.