Following a reference from "Elementos de Topología general" by Angel Tamariz and Fidel Casarrubias.
Since the sum and the product of real continous functions, defined in any topological space, are continuous we can easily prove that every metric space is a Tychonoff spaces. So we suppose that $(X,d)$ is a metric space and we consider a closed set $F$ a point $p\notin F$ and so we define a function $f:X\rightarrow[0,1]$ by the condiction
$$
f(x)=\frac{d(x,p)}{d(x,p)+d(x,F)}
$$
where $d(x,F):=\mathscr{inf}\{d(x,a):a\in F\}$. So we observe that $f(p)=0$ and $f(x)=1$ for any $x\in F$; then $f$ is continuous, since any function $g:X\rightarrow\mathbb{R}$ defined as
$$
g(x)=d(x,A)
$$for any $x\in X$ and for any fixed $A\subseteq X$, is a countinuous function and so $f$ is continuous function, since the sum and the ratio of continuous function is a continuous function.
Well by definition of Tychonoff space we can claim that the metric space $(X,d)$ is a Tychonoff space.
Well unfortunately I don't be able to prove that the function $g$ is continuous: could someone help me, please?
Best Answer
Not only $g$ is continuous, $g$ is, in fact, non-expansive: $$|g(x) - g(y)| \le d(x, y),$$ for all $x, y$. To prove this, suppose $x, y \in X$ and $a \in A$. Then, by triangle inequality, $$d(x, a) \le d(x, y) + d(y, a).$$ By definition of $d(x, A)$, we have $d(x, A) \le d(x, a)$, so, $$d(x, A) \le d(x, y) + d(y, a)$$ for all $x, y \in X$ and $a \in A$. Because this holds for all $a \in A$, we therefore see that $d(x, A) - d(x, y)$ is a lower bound for $d(y, a)$ when $a \in A$. So, the infimum must be larger than this lower bound, i.e. $$d(x, A) - d(x, y) \le \inf_{a \in A} d(y, a) = d(y, A).$$ Thus, $d(x, A) - d(y, A) \le d(x, y)$. Symmetrically, $d(y, A) - d(x, A) \le d(y, x) = d(x, y)$, hence $$|d(x, A) - d(y, A)| \le d(x, y),$$ as promised.