As you said, you need to prove there is an $N\in\mathbb{N}$ such that if $n\ge N$ then $d(f_n(x),f(x))<\epsilon$ for all $x\in K$. In other words, you want $f_n$ to be in $B^K_\varepsilon(f) = \{g:X\to Y \mid d(g(x),f(x))<\varepsilon \text{ for all } x\in K \}$. So if you find pairs $(K_1,V_1),\dots,(K_l,V_l)$ such that $f\in W=\bigcap_{i=1}^l(K_i,V_i)$ and $W\subseteq B^K_\varepsilon(f)$, then you can apply the convergence.
Your initial guess was to take $l=1$ and $K_1=K$, but this does not quite work. Consider for example $X=Y=K=[0,1]$ and $f=id$ and $\epsilon=0.5$. Then no open set $V$ exists such that $(K,V)$ is in $B_{0.5}(id_X)$.
Instead, note that for every $x\in K$ there is an open neighborhood $U_x$ of $x$ such that $f(U_x) \subseteq B_{\epsilon/3}(f(x)).$ Then
$$
f\left(\overline{U_x}\right) \subseteq \overline{f(U_x)} ⊆ \overline{B_{\epsilon/3}(f(x))} ⊆ B_{\epsilon/2}(f(x))
$$
Since $K$ is compact and the $U_x$ cover $K$, there are $x_1,\dots,x_l$ such that $U_i=U_{x_i},\ i=1,\dots,l$ cover $K$. Can you derive from these the pairs $(K_1,V_1),\dots,(K_l,V_l)$ such that each $g\in\bigcap_1^l(K_i,V_i)$ is an element of $B^K_\epsilon(f)$?
It's not implied by a lot of your properties as a countable discrete space lacks it, but such a space is second countable Hausdorff, separable etc.
It implies some other properties: $X$ has infinitely many open sets, all of which are infinite. (if $X$ would have finitely many open sets, their (open!) intersection would be minimal; If $U$ is open non-empty, define $U_0 = U$ and $\emptyset \neq U_{n+1} \subsetneq U_n$ whenever $X$ has this property, and then $\cup_n (U_{n}\setminus U_{n+1})$ is an infinite subset of $U$, so $U$ is infinite.
Spaces that are dense in themselves (i.e. have no isolated points) and are also $T_0$ will have this property: if $U$ is open then $U$ is not a singleton, so we have $x,y \in X$ with $x \neq y$, By $T_0$-ness we find an open set $V$ containing $x$ but not $y$ (or reversely), and then $U \cap V$ is strictly smaller and open too.
Equivalently, if $X$ fails the condition, there is some minimal open subset $U$. If $U$ has one point this point is isolated. If it has more, the $T_0$ condition fails for points from this set. So your condition is implied by "$X$ is dense in itself and $T_0$".
As @bof remarked in the comments: if $X$ has the "no minimal open set property", and $Y$ is any space whatsoever, then $X \times Y$ also obeys it ($O$ non-empty open in $X \times Y$ contains a basic non-empty open set $U \times V\subseteq O$ and $U$ is not minimal so $\emptyset \neq U' \subsetneq U$ open exists and then $U' \times V$ is smaller and non-empty open inside $O$ so $O$ is not minimal.)
Best Answer
I don't know if the property has a name, but it doesn't have to hold even for metric spaces. Consider $X=\mathbb{Q}$ with the Euclidean topology and let $A=\{0\}$. Clearly no open subset of $\mathbb{Q}$ is relatively compact (which is the same claim for metric, or even Hausdorff, spaces).
Proof. Note that for Hausdorff spaces the property "$V$ is contained in a compact set" is equivalent to "$\overline{V}$ is compact" which is also know as "$V$ is relatively compact".
"$\Rightarrow$" Since $\{x\}$ is compact then by our property it has open neighbourhood $U$ such that $\overline{U}$ is compact. Hence local compactness.
"$\Leftarrow$" Let $A\subseteq X$ be compact. Then for any $x\in A$ there is an open neighbourhood $U_x\subseteq X$ of $x$ that is relatively compact. Since $\{U_x\}_{x\in A}$ cover $A$ then by compactness $A$ is covered by $U_{x_1},\ldots,U_{x_n}$. Clearly $U_{x_1}\cup\cdots\cup U_{x_n}$ is the neighbourhood we are looking for. $\Box$
For non-Hausdorff spaces I suppose we can treat the property as one of the many non-equivalent definitions of local compactness. The name looks appropriate.