Existence of an arbitrary Small positive continuous real Valued Function Let $(X,\tau)$ be a Tychonoff Topological space. 
For each $x\in X$  consider an arbitrary  positive real number $\epsilon_x>0$. Is There a continuous real valued function $f:X\rightarrow \mathbb{R}$ with the following property:
$$\forall x \in X $$ 
$$0< f(x) < \epsilon_x$$

From the following comment of Edgar, I have Known that the following Question is the main Purpose of posing this Question, which I didn't notice to write it. 
Q.For which properties on $(X,\tau)$, we have the above Property? (one of the properties for which the above condition is true is that $(X, \tau)$ be discrete)
Statement: Is  the only property "discreteness" of $X$ ?
 A: The existence of a non-discrete T1 space with this property equivalent to the existence of a nonprincipal $\sigma$-complete ultrafilter (i.e. the existence of a measurable cardinal). Therefore, it is consistent with ZFC that all T1 spaces with that property are discrete.
First, suppose $\mathcal{U}$ is a nonprincipal $\sigma$-complete ultrafilter on the set $X$. Pick $\infty \notin X$ and define a topology on $X \cup \lbrace\infty\rbrace$ where all points of $X$ are isolated and the neighborhoods of $\infty$ are the sets $\lbrace\infty\rbrace\cup U$ where $U \in \mathcal{U}$. Suppose $\varepsilon_x \in (0,\infty)$ have been chosen for every $x \in X \cup \lbrace\infty\rbrace$. Since $\mathcal{U}$ is $\sigma$-complete, of the sets $$X_n = \lbrace x \in X : \varepsilon_x \gt 1/(n+1) \rbrace$$ eventually belong to $\mathcal{U}$. Let $n$ be such that $1/(n+1) \lt \varepsilon_\infty$ and $X_n \in \mathcal{U}$. Define $f:X\cup\lbrace\infty\rbrace\to(0,\infty)$ by $f(x) = 1/(n+1)$ when $x \in X_n \cup \lbrace\infty\rbrace$ and $f(x) = \varepsilon_x/2$ elsewhere. Then $f$ is continuous and $f(x) \lt \varepsilon_x$ for all $x \in X \cup\lbrace\infty\rbrace$.
For the converse implication, suppose $X$ is a space with the given property. 
First observe that the filter generated $\mathcal{N}$ by the neighborhoods of a point $x_0 \in X$ is $\sigma$-complete. To see that $\mathcal{N}$ is $\sigma$-complete, suppose $U_0 \supseteq U_1 \supseteq \cdots$ is a sequence of open neighborhoods of $x_0$ and let $Z = \bigcap_{n\lt\omega} U_n$. If $x \notin Z$ then define $\varepsilon_x = \min\lbrace 1/(n+1) : x \in U_n\rbrace$ and define $\varepsilon_x = 1$ on $Z$ (say). Suppose $f:X \to (0,\infty)$ is continuous and pick $n \geq 1$ so that $f(x_0) \geq 1/n$. Then there is an open neighborhood $U$ of $x_0$ such that $f(x) \gt 1/(n+1)$ for all $x \in U$. Thus $f(x) \gt 1/(n+1) \geq \varepsilon_x$ for any $x \in (U \cap U_n) - Z$. So if $f(x) \lt \varepsilon_x$ for every $x \in X$, then we must have $U \cap U_n \subseteq Z$, which shows that $Z$ contains an open neighborhood of $x_0$.
If $x_0$ is not an isolated point of $X$ and $X$ is T1 then $\mathcal{F} = \lbrace N-\lbrace x_0\rbrace: N \in \mathcal{N}\rbrace$ is a free filter on $X-\lbrace x_0\rbrace$ which is also $\sigma$-complete. This is not necessarily an ultrafilter, but I will show that there is a $Y \subseteq X-\lbrace x_0 \rbrace$ such that the restriction $\mathcal{F}|Y = \lbrace A \cap Y : A \in \mathcal{F}\rbrace$ is an ultrafilter, which is necessarily also $\sigma$-complete.
Indeed, suppose for the sake of contradiction that there is no such set $Y$, then we can find a countable partition of $X-\lbrace x_0 \rbrace$ into pairwise disjoint sets $X_n$ that are not in the ideal dual to $\mathcal{F}$. (Since $\mathcal{F}$ is not an ultrafilter, we can find sets $X_0, Y_0$ that are not in the dual ideal of $\mathcal{F}$ such that $X-\lbrace x_0 \rbrace = X_0 \cup Y_0$ and $X_0 \cap Y_0 = \varnothing$. Since $\mathcal{F}|Y_0$ is not an ultrafilter, we can similarly partition $Y_0 = X_1 \cup Y_1$. Repeat ad infinitum and throw any leftover points back into $X_0$.) Given such a partition, define $\epsilon_x = 1/(n+1)$ when $x \in X_n$ and $\epsilon_{x_0} = 1$. Suppose $f:X \to (0,\infty)$ is continuous and pick $n \geq 1$ so that $f(x_0) \geq 1/n$. Then there is a neighborhood $U$ of $x_0$ such that $f(x) \gt 1/(n+1)$ for all $x \in U$. Then $f(x) \gt 1/(n+1) = \epsilon_x$ for any $x \in U \cap X_n$. Since $X_n \cap U \neq \varnothing$, otherwise $X_n$ would be in the ideal dual to $\mathcal{F}$, we conclude that $f(x) \gt \varepsilon_x$ for some $x \in X$. Thus, we contradict the fact that our space has the given property.
A: The existence of a continuous function $f:X\rightarrow \mathbb{R} _ +$ with $ f(x)  <  \epsilon(x) $, for any given $\epsilon: X\rightarrow \mathbb{R} _ +$ is quite a strong assumption on $X$, as already observed in Gerald Edgar's and Andreas Blass' comments.
If we accept more reasonable assumptions on $\epsilon$, it is worth recalling the following simple but useful theorem by C.H.Dowker  (see Dugundji, Topology, VIII.4.3): 

If $X$ is paracompact and $\delta$ and $
> \epsilon  $ are real-valued functions
  on $X$,  $\delta$ upper semicontinuous
  and  $\epsilon$ lower
  semicontinuous, and if $\delta(x) <
> \epsilon (x) $ for all $x\in X$, then
  there exists a continuous $f$ on $X$
  with $\delta(x) < f(x) <  \epsilon(x)$
  on $X$.

A: We shall characterize those spaces in terms of a partition relation. This characterization is very similar to the property given in the question, but it may be useful and insightful.
Let $X$ be a Tychonoff space. Then the following are equivalent.


*

*Whenever $\epsilon_{x}>0$ there is some $x\in X$ there is some $f:X\rightarrow(0,\infty)$ with $f(x)>\epsilon_{x}$ for $x\in X$ (the proof is simpler if we replace $< $ with $>$).

*If $n_{x}\in\mathbb{N}$ for $n\in\mathbb{N}$, then there is a continuous mapping
$f:X\rightarrow\mathbb{N}$ such that $f(x)>n_{x}$ for $x\in X$. In other words, there are arbitrarily large continuous functions from $X$ to $\mathbb{N}$.

*If $P$ is a partition of $X$ into countably many sets, then there is some partition $Q$ of $X$ into clopen sets such that for each $B\in Q$ there are $A_{1},\dots,A_{n}\in P$ such that $B\subseteq A_{1}\cup\dots\cup A_{n}$.
$1\rightarrow 2$. Assume that if $\epsilon_{x}>0$ for $x\in X$ then there is a continuous mapping $f:X\rightarrow(0,\infty)$ with $f(x)<\epsilon_{x}$. Then as François G. Dorais showed, the neighborhood filter $\mathcal{N}(x)$ of every point $x\in X$ is $\sigma$-complete. Therefore the space $X$ is a $P$-space. It is well known and one can easily show that a completely regular space is a $P$-space if and only if whenever $f:X\rightarrow\mathbb{R}$ is continuous, then around each point $x\in X$ there is a neighborhood $U$ of $x$ with $f''(U)=\{f(x)\}$. In other words, $P$-spaces are precisely the spaces where every continuous real-valued function is locally constant.
Now assume that $n_{x}\in\mathbb{N}$ for $x\in X$. Then there is some function $f:X\rightarrow\mathbb{R}$ such that $f(x)>n_{x}$ for $x\in X$. Let $\mathbb{R}^{d}$ be the real numbers with the discrete topology. Then since $X$ is a $P$-space, the function $f$ is locally constant, so $f$ is a continuous function from $X$ to $\mathbb{R}^{d}$. Let $g:\mathbb{R}^{d}\rightarrow\mathbb{N}$ be a function with $g(x)\geq x$ for $x\in X$. Then we have $g\circ f:X\rightarrow\mathbb{N}$ be a continuous function with $g\circ f(x)\geq f(x)>n_{x}$ for $x\in X$. 
$2\rightarrow 1$ This is obvious.
$2\rightarrow 3$. Assume that $P=\{A_{1},\dots,A_{n},\dots\}$ is a partition of $X$ into countably many sets. Then if $x\in A_{n}$, then assume that $n_{x}=n$. Then there is a continuous $f:X\rightarrow\mathbb{N}$ such that $f(x)>n_{x}$ for all $x\in X$. Let $B_{n}=f_{-1}(\{n\})$ for all $n$. We claim that $B_{n}\subseteq A_{1}\cup\dots\cup A_{n}$. If $x\in B_{n}$, then $n_{x}< f(x)=n$, so $x\in A_{n_{x}}$ for some $n_{x}< n$, so $x\in A_{1}\cup\dots\cup A_{n}$. Thus $B_{n}\subseteq A_{1}\cup....\cup A_{n}$.
$3\rightarrow 2$. Assume that whenever $P$ is a countably partition of $X$, then there is countable partition $Q$ of $X$ into clopen sets where for each $B\in Q$ there are
$A_{1},\dots,A_{n}\in P$ with $B\subseteq A_{1},\dots,A_{n}$. Now assume that $n_{x}\in\mathbb{N}$ for $x\in X$. Then let $A_{n}=\{x\in X|n_{x}=n\}$ for all $n$. Then there is a partition $Q=\{B_{1},\dots,B_{n},\dots\}$ of $X$ into clopen sets such that for all $n$ there is a function $g:\mathbb{N}\rightarrow\mathbb{N}$ such that $B_{n}\subseteq A_{1}\cup\dots\cup A_{g(n)}$ for all $n$. Let $f:X\rightarrow\mathbb{N}$ be the function where if $x\in B_{n}$, then $f(x)=g(n)+1$. Then since $x\in B_{n}\subseteq A_{1}\cup\dots\cup A_{g(n)}$, we have $x\in A_{i}$ for some $i\leq g(n)$, so $n_x=i\leq g(n)< g(n)+1=f(x)$. Furthermore, since each $B_{n}$ is clopen, we have $f$ be a continuous function.
A: So you are asking: which topological spaces, besides the discrete ones, 
are such that for every strictly positive real function $g$ there is a 
strictly positive continuous real function $f$ with $ 0 < f < g $? Only some hints.
Others have already noted that there cannot be nontrivial convergent sequences.
You can note that if there are no nontrivial convergent sequences, then replace $g$
with the largest $h<g$ which takes only values of the form $1/n$, and then note that
this $h$, even if not continuous, at least does not give the above noted problem
(forcing a possible continuous $f$ to have value $0$). Taking the closures of the sets
where $h>1/n$ one can then use Urhyson's lemma / Tietze extension theorem (when the space
is normal) and so obtain a continuous $f$ with $0<f<h$.
Is there a non-discrete normal space with no nontrivial converging sequence?
You can easily find the answer in any standard book on general topology which
treats Stone compactification.
Added: please replace "no nontrivial convergent sequences" with the stronger
"every countable subset is closed". And examples are even more exotic, but Gillman
and Jerison, rings of continuous functions, should have them (and perhaps also
Engelking).
Concerning TeX commands: I do not use them on purpose, but if it this the rule
to always use them here, then I will in future only give answers that do not require 
mathematical notation, so that we all will be able to read in the way we like.
