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).

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 such that $X_n \in \mathcal{U}$ for every $m \geq n$. 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 by the neighborhoods of a point in $X$ is $\sigma$-complete. Let $\mathcal{N}$ be the filter on $X$ generated by the neighborhoods of $x_0 \in X$. 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 $x \in U$. Then $f(x) \gt 1/(n+1) \geq \varepsilon_x$ 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 an $Y \subseteq X-\lbrace x_0 \rbrace$ 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 $X_n \cap U \neq \varnothing$, 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.

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.