Suppose the countable subspace $D$ is dense in the separable Tychonoff space $X$ and $f$ is a continous function from $D$ to the closed unit interval. What are some conditions on $X$ or $D$, which make $f$ continuously extendable over $X$？

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), 204206 (1951)] is easily proved, as well as a theorem of Vulih [Mat. Sbornik N.S. 30(72), 167170 (1952); MR0048790 (14,70c)]. A final corollary is a special case of a theorem widely known and recently published by Katětov [Fund. Math. 38, 8591 (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). 


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. 


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. 


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. 

