Let $X$ be a completely regular and hausdorff space. For a continuous function $f :X\longrightarrow \Bbb{R}$ define $A_f := \{ (x, \frac 1{f(x)}) \; | \; f(x)\not = 0 \}$. Assume that for any continuous function $g: X \longrightarrow \Bbb{R}$, $\mbox{Graph}(g)\cap A_f$ is finite. Does it mean that $A_f$ is finite?

New version of the problem I am looking for a characterization of those completely regular and hausdorff spaces $X$ such that the follwing is true:

If $f :X\longrightarrow \Bbb{R}$ is continuous such that for any continuous function $g: X \longrightarrow \Bbb{R}$, $\mbox{Graph}(g)\cap A_f$ is finite, then $A_f$ is finite. (where $A_f := \{ (x, \frac 1{f(x)}) \; | \; f(x)\not = 0 \}$).

As @NikWeaver has pointed out below this is now true in general.