Extending continuous function $D\to I$, where $D$ is a dense subspace of a separable Tychonoff space Suppose the countable subspace $D$ is dense in the separable Tychonoff space $X$ and $f$ is a continuous function from $D$ to the closed unit interval. What are some conditions on $X$ or $D$, which make $f$ continuously extendable over $X$？
 A: A relevant paper is 
Taĭmanov, A. D.
On extension of continuous mappings of topological spaces. (Russian)
Mat. Sbornik N.S. 31(73), (1952). 459–463.
56.0X
The MR of this paper is:
Let $S$ be a $T_1$-space, $A$ a dense subspace of $S$, and $R$ a compact Hausdorff space. Let $f$ be a continuous mapping of $A$ into $R$. Then $f$ admits a continuous extension over $S$ if and only if for all disjoint closed subsets $A_1,A_2$ of $R$, the relation $(f^{-1}(A_1))^-\cap(f^{-1}(A_2))^-=0$ obtains (closure in $S$). From this result, a theorem of Yu. M. Smirnov [Uspehi Matem. Nauk 6, no. 4(44), 204--206 (1951)] is easily proved, as well as a theorem of Vulih [Mat. Sbornik N.S. 30(72), 167--170 (1952); MR0048790 (14,70c)]. A final corollary is a special case of a theorem widely known and recently published by Katětov [Fund. Math. 38, 85--91 (1951); MR0050264 (14,304a)]. 
I remembered this because as a graduate student I used it to give a (I think new at the time) proof that every compact Hausdorff space is a continuous image of a compact totally disconnected compact Hausdorff space (which, in turn, I use these days to reduce proving  the Riesz representation theorem for $C(K)$ to the case where the compact space $K$ is totally disconnected).
A: The criterion for "EVERY continuous map from $D$ to $[0, 1]$ has a continuous extension to $X$" is that any two disjoint zerosets in $D$ have disjoint closures in $X$. You can find this in Chapter 6 of Gillman and Jerison's classic "Rings of Continuous Functions".  They also consider the "local problem" of continuously extending a single map at length in some of the exercises, e.g. given $f:D\rightarrow Y$ (not necessarily $Y=[0, 1]$) Exercise 6G characterizes the largest subspace of $X$ to which $f$ can be continuously extended in terms of $z$-filters.
A: The situation is analogous to the particular case of $X$ a metric space, for any Tychonoff space $X$ is uniformisable, and a real valued function $f$ on a dense subset $D$ of a uniform space $X$ is certainly continuously extendable to $X$ provided it is uniformly continuous. This is also a necessary condition if $X$ is compact, for any continuous function on a compact uniform space is always uniformly continuous.
A: At least in the case of a metric space $X$, such a function $f$ extends from $D$ to all of $X$ if and only if $f$ maps Cauchy sequences to Cauchy sequences (note that this is a weaker condition than uniform continuity).
As mentioned by Pietro, for your general Tychonoff space $X$, you make it a uniform space, so I think you can generalize my statement above to the following: $f$ extends if and only if $f$ maps Cauchy nets to Cauchy nets.
A: Several years ago I encountered a similar problem in my joint investigation with Misha Mitrofanov and I solved it as follows (see our paper “Approximation of continuous functions on Fréchet spaces” (English version is downloadable in the source file)).
We say that a filter $\mathcal F$ on a topological space is convergent, if a set of limits of $\mathcal F$ is non-empty. 
Let $X$  be a topological space, $Y$ be a regular topological space, $D$  be a dense subset of the space $X$,  $g : D\to Y$  be a map, and let  $\mathcal  F$ be a filter on the space  $X$.  Let  $g(F)$ be a filter on the space  $Y$  generated by the base $\{g(F): F\in \mathcal F\}$.  For every point  $x\in X$,  let $\mathcal F_x$ be the trace of the filter of all neighborhoods of the point $x$ on the set $D$. 
Lemma 1.  A continuous map $f : D\to Y$ can be extended to a continuous map
 $\tilde f : X\to Y$ iff for any point  $x\in X$ the filter $f(\mathcal F_x)$  is convergent.
We can simplify this condition when $X$ is a Fréchet–Urysohn space. That is if for any $A\subset X $ and any points $x\in\overline{A}$ there exists a sequence of points of the set $A$  convergent to  $x$.  In particular, every first countable space (and, hence, every metrizable space) is a Fréchet–Urysohn space. 
Lemma 2.  Let  $X$   be a Fréchet-Urysohn topological space, $Y$  be a regular topological space, and $D$  be a dense subset of the space $X$.  A continuous mapping   $f : D\to Y$  can be extended to a continuous mapping $\hat f : X\to Y$ iff for each sequence $\{x_n\}$ of points of the set $D$ convergent in $X$, a sequence  $\{ f(x_n)\}$ is also convergent.
