This is not really an answer but too long for a comment:

Let me point out that there might be a relation to the Alexander Subbase Theorem here:
A topological space is compact iff the topology has a subbase with the FSP.

What FIP gives you is this:

If $R$ has the FSP and is a subbase for the topology of a space $X=\bigcup R$, then $X$ has a basis (namely all finite intersections of elements of $R$) with the FSP.

Unfortunately it seems that getting compactness from the existence of a basis with the FSP also requires some form of choice. So FIP could be strictly weaker than the Alexander Subbase Theorem. But an upper bound on how much choice is needed for the subbase theorem would also be an upper bound for FIP.

Unfortunately I don't have access to any Axiom of Choice book. But it seems likely that there is some information about the amount of choice needed for the subbasis theorem.

Edit: By godelian's comment below, this actually answers the question: FIP follows from Alexander's Subbase Theorem which follows from the Boolean prime ideal theorem which is known to be strictly weaker than the full Axiom of Choice. So no, FIP does not imply AC.

This is not really an answer but too long for a comment:

Let me point out that there might be a relation to the Alexander Subbase Theorem here:
A topological space is compact iff the topology has a subbase with the FSP.

What FIP gives you is this:

If $R$ has the FSP and is a subbase for the topology of a space $X=\bigcup R$, then $X$ has a basis (namely all finite intersections of elements of $R$) with the FSP.

Unfortunately it seems that getting compactness from the existence of a basis with the FSP also requires some form of choice. So FIP could be strictly weaker than the Alexander Subbase Theorem. But an upper bound on how much choice is needed for the subbase theorem would also be an upper bound for FIP.

Unfortunately I don't have access to any Axiom of Choice book right now. But it seems likely that there is some information about the amount of choice needed for the subbasis theorem.

Edit: By godelian's comment below, this actually answers the question: FIP follows from Alexander's Subbase Theorem which follows from the Boolean prime ideal theorem which is known to be strictly weaker than the full Axiom of Choice. So no, FIP does not imply AC.

This is not really an answer but too long for a comment:

Let me point out that there might be a relation to the Alexander Subbase Theorem here:
A topological space is compact iff the topology has a subbase with the FSP.

What FIP gives you is this:

If $R$ has the FSP and is a subbase for the topology of a space $X=\bigcup R$, then $X$ has a basis (namely all finite intersections of elements of $R$) with the FSP.

Unfortunately it seems that getting compactness from the existence of a basis with the FSP also requires some form of choice. So FIP could be strictly weaker than the Alexander Subbase Theorem. But an upper bound on how much choice is needed for the subbase theorem would also be an upper bound for FIP.

Unfortunately I don't have access to any Axiom of Choice book. But it seems likely that there is some information about the amount of choice needed for the subbasis theorem.

This is not really an answer but too long for a comment:

Let me point out that there might be a relation to the Alexander Subbase Theorem here:
A topological space is compact iff the topology has a subbase with the FSP.

What FIP gives you is this:

If $R$ has the FSP and is a subbase for the topology of a space $X=\bigcup R$, then $X$ has a basis (namely all finite intersections of elements of $R$) with the FSP.

Unfortunately it seems that getting compactness from the existence of a basis with the FSP also requires some form of choice. So FIP could be strictly weaker than the Alexander Subbase Theorem. But an upper bound on how much choice is needed for the subbase theorem would also be an upper bound for FIP.

Unfortunately I don't have access to any Axiom of Choice book. But it seems likely that there is some information about the amount of choice needed for the subbasis theorem.

Edit: By godelian's comment below, this actually answers the question: FIP follows from Alexander's Subbase Theorem which follows from the Boolean prime ideal theorem which is known to be strictly weaker than the full Axiom of Choice. So no, FIP does not imply AC.