Expressive adequacy After reading a few others posts on this site, I am still struggling in showing how sets are incomplete.
(a) Show that {$↔,\bot, ∧$} is functionally complete, but that no proper subset is.
(b) is solved! :)
(b) Assume that $c$ is a 2-place connective. Show that if either $f_c(\top, \top) = \top$ or $f_c(\bot, \bot) = \bot$, then {$c$} is not complete. 
For (b), I understand that for {$\uparrow$} and {$\downarrow$}, the only functionally complete 2-place connectives, that $f_{\uparrow}(\top, \top) = \bot$ and $f_{\uparrow}(\bot, \bot) = \top$. The same results apply for $f_{\downarrow}$. I just can't see how to then apply this to prove $f_c(\top, \top) = \top$ means {$c$} is incomplete.

As Noah points out in the comments, you know that $c$ cannot be either the  $\uparrow$ and $\downarrow$, and since those are the only two connectives that are by themselves complete, you know $c$ is not complete.
However, I doubt that this proof is 'acceptable', or at least probably not what was expected of you. Probably you had to give a more direct proof why such a $c$ is not complete. Well, think about it. Suppose you have any number of connectives $c$, applied to any number of instances of $\top$, e.g. $(\top c \top) c (\top c (\top c \top))$ ... if you evaluate this, you'll of course get $\top$ for any expression; you';ll never get $\bot$.
Now, to make that idea into a rigorous proof, use induction: prove by (strong) induction on the number of operators $c$ in any statement $\phi$ that is composed of $c$'s and $\top$, that $\phi$  will always evaluate to $\top$. I'll leave the details to you.
