1) $PC(X,Y)$ is the topology itself. The sets $[x_1,\ldots,x_n;O_1,\ldots,O_n]$ form a base, by definition.
These sets are closed under intersections:
$$[x_1,\ldots,x_n;O_1,\ldots,O_n] \cap [x'_1,\ldots,x'_m;O'_1,\ldots,O'_m] = [x_1,\ldots,x_n,x'_1,\dots,x'_m; O_1,\ldots,O_n,O'_1,\ldots,O'_m]$$
and they cover $C(X,Y)$, clearly, so they form the base for some topology.
Note that this is just the subspace topology $C(X,Y)$ as a subspace of the product topology on $Y^X = \prod_{x \in X} Y$, all functions (not necessarily continuous) from $X$ to $Y$. This holds because these sets are basically just product sets that only depend on finitely many coordinates (the $x_i$) and have no limitations on the other coordinates.
2) Yes, these topologies are equivalent. The sets $[A;V]$ also form a base for $PC(X,Y)$ from 1. They are certainly open: if $A = \{x_1,\ldots,x_n\}$ then $[A;V] = [x_1,\ldots,x_n;V,\ldots,V]$. And $$[x_1,\ldots,x_n;O_1,\ldots,O_n] = \cap_{i=1}^n [\{x_i\}; O_i]$$ so the other basic elements are open in this topology too. So the basic sets are "mutually open" so the topologies coincide.
That $x$ is a limit point of $A$ implies that it is a limit point of $A\setminus F$, where $F \subseteq A$ is finite, will hold in any $T_1$ space (so when singletons are closed). That answers the first part of your question.
In a $T_1$ space, being a limit point of $A$ (every neighbourhood of $x$ intersects $A\setminus\{x\}$) is equivalent to $x$ being an $\omega$-limit point of $A$ (every neighbourhood of $x$ contains infinitely many points of $A$).
"Every infinite set $A$ has an $\omega$-limit point" is in general spaces equivalent to "every countable open cover of $X$ has a finite subcover", i.e. countable compactness, which for metrisable spaces indeed is equivalent to compactness but in general not: consider $\omega_1$, the first uncountable ordinal in the order topology, or $\{0,1\}^\mathbb{R}\setminus \{\underline{0}\}$, which are countably compact but not compact. The first is also sequentially compact, first countable, hereditarily normal etc., so even for very nice spaces these notions need not be equivalent.
Best Answer
The union is trivial: if $U_i, i \in I$ are all open and $f \in U=\bigcup_i U_i$, then for some $j \in I$, $f \in U_j$. By definition we then have $x_1,\ldots x_n$ in $X$ and $O_1,\ldots,O_n$ open in $Y$ such that $f \in [x_1,\ldots,x_n; O_1, O_2,\ldots O_n] \subseteq U_i \subseteq U$.
For finite intersections we only need to check the intersection of two open sets $U_1$ and $U_2$ (finite then follows by induction, a standard argument), so let $f \in U_1 \cap U_2$. Then $f \in U_1$ gives us $x_1,\ldots x_n$ in $X$ and $O_1,\ldots,O_n$ open in $Y$ such that $f \in [x_1,\ldots,x_n; O_1, O_2,\ldots O_n] \subseteq U_1$ and $f \in U_2$ gives us $x'_1,\ldots x'_m$ in $X$ and $O'_1,\ldots,O'_m$ open in $Y$ such that $f \in [x'_1,\ldots,x'_m; O'_1, O'_2,\ldots O'_m] \subseteq U_2$. But then
$$f \in [x_1,\ldots,x_n, x'_1,\ldots, x'_m; O_1,\ldots,O_n, O'_1,\ldots,O'_m] \subseteq \\ [x_1,\ldots,x_n; O_1, O_2,\ldots O_n] \cap [x'_1,\ldots,x'_m; O'_1, O'_2,\ldots O'_m] \subseteq U_1 \cap U_2$$
as required.
You can also note that basic open neighbourhood of $f$ (the $[x_1,\ldots,x_n; O_1, \ldots, O_n]$) is of the form $B \cap C(X,Y)$ where $B$ is a basic open subset of $Y^X$ in the product topology, given by a finite subset $\{x_1, \ldots,x_n\}$ of "coordinates" from $X$ and finitely many $O_1, \ldots, O_n$ in those coordinates. So e.g. $f_n \to f$ in this topology iff $f_n(x) \to f(x)$ for all $x \in X$, as this holds for the product topology (and likewise for nets), hence the name "pointwise topology" for this space $C_p(X,Y)$. The "$p$" could also be seen as standing for "product", as it's just the product topology on the set, essentially.