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.