Topology generated by Functions. Fix a set $X$. Let $\mathcal{F}_0$ be a set of functions $g:X\to\mathbb{R}$. Let $\mathcal{T}_0$ be the smallest topology on $X$ in which all $g\in\mathcal{F}_0$ are continuous. Next, we say that a function $f:X\to\mathbb{R}$ is locally generated by $\mathcal{F}_0$, if for every $x\in X$, there exists an open neighborhood $U\in\mathcal{T}_0$ of $x$, $k\in\mathbb{N}$, $g_1,...,g_k\in\mathcal{F}_0$, and a continuous map $F:\mathbb{R}^k\to\mathbb{R}$ such that forall $y\in U$, we have $f(x)=F(g_1(x),...,g_k(x))$. Let $\mathcal{F}$ be the set of all functions $f:X\to\mathbb{R}$ locally generated by $\mathcal{F}_0$, and $\mathcal{T}$ the smallest topology in which all $f\in\mathcal{F}$ are continuous.
Note that one can prove $\mathcal{T}_0=\mathcal{T}$.
Now assuming that $\mathcal{T}$ is $T_0$ topology, prove directly that it is $T_3$ (regular), and then prove that it is $T_{3\frac{1}{2}}$ (completely regular).
Can somebody help me with proving this?? I can't even start...
Also, if anybody have any reference related to this problem, let me know then I will appreciate it. Thank you in advance...

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On the other hand, I tried to disprove the above problem by giving a counter-example... If any one interested in the following claim, you may check and verify. Furthermore, if you can find any mistake or wrongness on the claim, and if you let me know, I will appreciate so much... Thank you.
Now, I thought the following example was a counter example, but I got $0$ score on this.. (It was a homework whether to prove or disprove (by giving a counter example).)
Let $X=\mathbb{R}$ with $\mathcal{F}_0=\{f,g\}$ where $f$ is defined by $f(x)=x$ for $x\geq 0$, and $f(x)=-\frac{1}{2}x$ for $x<0$, and $g(x)=-\frac{1}{2}x$ for $x\geq 0$ and $g(x)=x$ for $x<0$. Then, by drawing a graph, you can easily see that the smallest topology generated by this functions is $T_0$ space (You can choose arbitrary small open interval on $Y$-axis to separte a point (in $X$) from another point (in $X$)). Now consider a point $-\frac{1}{2}$ and a closed subset $[0,1]$. Since for every $(a,b)\subset\mathbb{R}$ such that $\frac{1}{2},\frac{1}{4}\in(a,b)$, either $f^{-1}(a,b)$ or $g^{-1}(a,b)$ intersects $[0,1]$ while the $f^{-1}(a,b)$ or $g^{-1}(a,b)$ are basic open sets in $\mathcal{T}_0$, that containing $-\frac{1}{2}\in\mathbb{R}$, I claim that it is not regular, and hence not completely regular.
 A: It seems the following. 
A map $G:X\to\mathbb R^{{\mathcal F}_0}$ such that $x\mapsto (f(x)_{f\in\mathcal F})$ is a diagonal product of a family of continuous maps and hence $G$ is continuous too. Since the space $(X,\mathcal{T}_0)$ is $T_0$, the map $G$ is injective. Let $\mathcal{T}_1$ be the preimage under the map $G$ of the topology of the subspace $G(X)$ of the space $\mathbb R^{{\mathcal F}_0}$. Since the map $G$ is continuous, we see that $\mathcal{T}_1\subset \mathcal{T}_0$. From the other side, let $f\in {\mathcal F}_0$ be an arbitrary function. Then $f=\pi_f\circ G$, where $\pi_f$ is the projection of the space $\mathbb R^{{\mathcal F}_0}$ onto its $f$-th coordinate. So $f$ is a continuous map on a space $(X,\mathcal{T}_1)$. The minimality of the topology $\mathcal{T}_0$ implies $\mathcal{T}_0\subset \mathcal{T}_1$. Hence $\mathcal{T}_1=\mathcal{T}_0$ and  $G(X)$ is a homeomorphic image of $X$. The space $G(X)$ is completely regular as a subspace of a completely regular space $\mathbb R^{{\mathcal F}_0}$.


On the other hand, I tried to disprove the above problem by giving a counter-example... If any one interested in the following claim, you may check and verify. Furthermore, if you can find any mistake or wrongness on the claim, and if you let me know, I will appreciate so much... 
Let $X=\mathbb{R}$ with $\mathcal{F}_0=\{f,g\}$ where $f$ is defined by $f(x)=x$ for $x\geq 0$, and $f(x)=-\frac{1}{2}x$ for $x<0$, and $g(x)=-\frac{1}{2}x$ for $x\geq 0$ and $g(x)=x$ for $x<0$.

It seems that the smallest topology $\mathcal{T}_0$  generated by the set $\mathcal{F}_0$ is just the standard topology of $\mathbb R$. Indeed, let $x>0$ and $0<\varepsilon<3x/5$. Then  
$$x\in f^{-1}(x-\varepsilon,  x+\varepsilon)\cap g^{-1}(x-\varepsilon,  x+\varepsilon)=
\left(\left(-(x+\varepsilon)/2, -(x-\varepsilon)/2\right)\cup\left(x-\varepsilon, x+\varepsilon\right) \right)\cap
\left(\left(-2x-2\varepsilon, -2x+2\varepsilon\right)\cup\left(x-\varepsilon, x+\varepsilon\right)\right)=\left(x-\varepsilon, x+\varepsilon\right).$$
The case $x<0$ is considered similarly. 
