Daniil wrote an excellent post, but just to add to that a little bit:
As Daniil pointed out, you can't capture any truth-functions that non-trivially depend on more than $1$ variable, such as $P \land Q$, with only a $\neg$. So, let's restrict ourselves to functions defined over one variable, $P$, and see if maybe we can capture all those using a $\neg$?
Unfortunately, the answer is still no. Again, as Daniil already pointed out, we can't capture any tautology or contradiction. That is, we can't capture the truth-function that always returns true (i.e. the function $f$ such that $f(T)=f(F)=T$), nor can we capture the truth-function that always returns false (i.e. the function $f'$ such that $f'(T)=f'(F)=F$)
So in this post I just wanted to show you how you can prove that result using induction. In particular, let's prove the following:
Claim
For any expression $\phi$ built up from $P$ and $\neg$ alone, it will be true that if $v$ is the valuation that sets $P$ to true (i.e. $v(P)=T$), and $v'$ is the valuation that sets $P$ to false (i.e. $v'(P)=F$), then either $v(\phi)=T$ and $v'(\phi)=F$, or $v'(\phi)=T$ and $v(\phi)=F$ (in other words, $v(\phi)$ and $v'(\phi)$ will always opposite values, meaning that $\phi$ can not be a tautology or contradiction, for that would require that $\phi$ has the same value for any valuation)
Proof
We'll prove the claim by structural induction on the formation of $\phi$:
*Base: *
$\phi=P$. Then $v(\phi)=v(P)=T$, while $v'(\phi)=v'(P)=F$. Check!
Step:
If $\phi$ is not an atomic proposition, then there is only one possibility: $\phi$ is the negation of some other statement $\psi$, i.e. $\phi = \neg \psi$.
Now, by inductive hypothesis we can assume that $v(\psi)=T$ and $v'(\psi)=F$, or $v'(\psi)=T$ and $v(\psi)=F$
Well, if $v(\psi)=T$ and $v'(\psi)=F$, then $v(\phi)=v(\neg \psi)=F$ and $v'(\phi)=v'(\neg \psi) =T$. On the other hand, if $v(\psi)=F$ and $v'(\psi)=T$, then $v(\phi)=v(\neg \psi)=T$ and $v'(\phi)=v'(\neg \psi) =F$. So, we can conclude that $v(\phi)=T$ and $v'(\phi)=F$, or $v'(\phi)=T$ and $v(\phi)=F$, as desired.
Sorry to be brief, but here are some things one would want to check for right away: Soundness, Completeness, and as you mentioned, Compactness.
Have a look at other logics that are close to First-Order Logic. The first thing that comes to mind is Intuitionistic Logic, but there are others, for instance modal logics.
Best Answer
Unless I'm missing something, the problem is incorrect: $\neg a$ is equivalent to $a\iff\perp$. Since you have $\vee$ included explicitly, as you observe the functional completeness of $\{\vee,\neg\}$ translates to your set of connectives.