It seems the following.

Dealing with continuous functions on a topological space $X$, it is natural to consider $X$ to be Tychonoff, or, at least, functionally Hausdorff. I recall that a space $X$ is called functionally Hausdorff provided for each distinct points $x,y\in X$ there exists a continuous function $f:X\to[0;1]$ such that $f(x)=0$ and $f(y)=1$. Clearly, each functionally Hausdorff space is Hausdorff.

Proposition. Each functionally Hausdorff space $X$ possessing the property is finite.

Proof. Assume the converse. By induction we construct a sequence $\{a_n\}$ of points of the space $X$ and a sequence $\{f_n\}$ of continuous functions from the space $X$ into the segment $[0;1]$ as follows. Let $a_0’$ and $a_0''$ be arbitrary distinct points of the space $X$. The functional Hausdorfness of the space $X$ implies that there exists a continuous function $f_0’$ from the space $X$ into the segment $[0;1]$ such that $f_0’(a_0’)=0$ and $f_0’(a_0’’)=1$. If the set $f_0’^{-1}([1/2;1])$ is infinite, put $a_0=a_0’$ and $f_0=f’_0$, in the opposite case we put put $a_0=a_0’’$ and $f_0=1-f’_0$. In both cases $f_0(a_0)=0$ and a set $f_0^{-1}([1/2;1])$ is infinite. Let $a_1’$ and $a_1’’$ be arbitrary distinct points of the set $f_0^{-1}([1/2;1])$. The functional Hausdorfness of the space $X$ implies that there exists a continuous function $f_1’$ from the space $X$ into the segment $[0;1]$ such that $f_1’(a_1’)=0$ and $f_1’(a_1’’)=1$. If the set $ f_0^{-1}([1/2;1])\cap f_1'^{-1}([1/2;1])$ is infinite, put $a_1=a_1’$ and $f_1=f’_1$, in the opposite case we put put $a_1=a_1’’$ and $f_1=1-f’_1$. In both cases $f_1(a_1)=0$ and a set $f_0^{-1}([1/2;1])\cap f_1^{-1}([1/2;1])$ is infinite. Let $a_2’$ and $a_2’’$ be arbitrary distinct points of the set $f_0^{-1}([1/2;1])\cap f_1^{-1}([1/2;1])$, and so forth.

Put $A=\{a_n\}$. Now let $a_n\in A$ be an arbitrary point. Let $g:\Bbb R\to\Bbb R$ be a continuous function sucn that $g(0)=1$ and $g(t)=0$ for each $t\ge 1/2$. Define a (continuous) function $g:X\to\Bbb R$ by putting for each point $x\in X$ $$f(x)=(g\circ f_n(x))\cdot f_1(x)\cdots f_{n-1}(x).$$ Then $f(x)=0$ for each point $x(A\setminus\{a_n\} )$, but $f(a_n)\ge 1/2^{n-1}$, a contradiction. $\square$

If you are interested in the space $X$ with weaker separation axioms than functional Hausdorffness, then this recent MSE question seems to be a Hausdorff counterpart of yours.

It seems the following.

Dealing with continuous functions on a topological space $X$, it is natural to consider $X$ to be Tychonoff, or, at least, functionally Hausdorff. I recall that a space $X$ is called functionally Hausdorff provided for each distinct points $x,y\in X$ there exists a continuous function $f:X\to[0;1]$ such that $f(x)=0$ and $f(y)=1$. Clearly, each functionally Hausdorff space is Hausdorff.

Proposition. Each functionally Hausdorff space $X$ possessing the property is finite.

Proof. Assume the converse. By induction we construct a sequence $\{a_n\}$ of points of the space $X$ and a sequence $\{f_n\}$ of continuous functions from the space $X$ into the segment $[0;1]$ as follows. Let $a_0’$ and $a_0''$ be arbitrary distinct points of the space $X$. The functional Hausdorfness of the space $X$ implies that there exists a continuous function $f_0’$ from the space $X$ into the segment $[0;1]$ such that $f_0’(a_0’)=0$ and $f_0’(a_0’’)=1$. If the set $f_0’^{-1}([1/2;1])$ is infinite, put $a_0=a_0’$ and $f_0=f’_0$, in the opposite case we put put $a_0=a_0’’$ and $f_0=1-f’_0$. In both cases $f_0(a_0)=0$ and a set $f_0^{-1}([1/2;1])$ is infinite. Let $a_1’$ and $a_1’’$ be arbitrary distinct points of the set $f_0^{-1}([1/2;1])$. The functional Hausdorfness of the space $X$ implies that there exists a continuous function $f_1’$ from the space $X$ into the segment $[0;1]$ such that $f_1’(a_1’)=0$ and $f_1’(a_1’’)=1$. If the set $ f_0^{-1}([1/2;1])\cap f_1'^{-1}([1/2;1])$ is infinite, put $a_1=a_1’$ and $f_1=f’_1$, in the opposite case we put put $a_1=a_1’’$ and $f_1=1-f’_1$. In both cases $f_1(a_1)=0$ and a set $f_0^{-1}([1/2;1])\cap f_1^{-1}([1/2;1])$ is infinite. Let $a_2’$ and $a_2’’$ be arbitrary distinct points of the set $f_0^{-1}([1/2;1])\cap f_1^{-1}([1/2;1])$, and so forth.

Put $A=\{a_n\}$. Now let $a_n\in A$ be an arbitrary point. Let $g:\Bbb R\to\Bbb R$ be a continuous function sucn that $g(0)=1$ and $g(t)=0$ for each $t\ge 1/2$. Define a (continuous) function $g:X\to\Bbb R$ by putting for each point $x\in X$ $$f(x)=(g\circ f_n(x))\cdot f_1(x)\cdots f_{n-1}(x).$$ Then $f(x)=0$ for each point $x(A\setminus\{a_n\} )$, but $f(a_n)\ge 1/2^{n-1}$, a contradiction. $\square$

If you are interested in the space $X$ with weaker separation axioms than functional Hausdorffness, then this recent MSE question seems to be a Hausdorff counterpart of yours.