Firstly, I think, "$R_1$" is more common than "preregular". Therefore, I prefer to use $R_1$, if you don't mind.
STEP 1: This holds for $T_2$. More generally, let $(A_i)_{i \in I}$ be a locally finite cover of $X$, such that each $A_i$ is closed and $T_2$. Then $X$ is $T_2$.
PROOF. Let $x, y$ be distinct elements of $X$.
Define $J := \{i \in I: x \in A_i \}$ and
$K := \{i \in I: y \in A_i \}$.
Note that $J$ and $K$ are finite.
For $i \in J \cap K$ there exist $U_i, V_i$ open in $X$, such that
$x \in U_i, y \in V_i, \space U_i \cap V_i \cap A_i = \emptyset$.
$U := \bigcap_{i \in J\cap K} U_i \space\cap\space (X \setminus \bigcup_{i \in I \setminus J} A_i)$
$V := \bigcap_{i \in J\cap K} V_i \space\cap\space (X \setminus \bigcup_{i \in I \setminus K} A_i)$
By local finiteness, $U, V$ are open.
It is easy to see that $x \in U, y \in V$ and $U \cap V = \emptyset$.
NOTE: In fact, I found this version easier to prove rather than to consider all the different cases in the original question, where the cover consists of only two elements.
STEP 2: Let $(A_i)_{i \in I}$ be a locally finite cover of $X$, such that each $A_i$ is closed and $R_1$. Then $X$ is $R_1$.
PROOF.
Consider the Kolmogorov quotient $KQ(X)$ of $X$ and the canonical map $\lambda: X \rightarrow KQ(X)$.
It is easy to see that $X$ is $R_1$, if and only if $KQ(X)$ is $T_2$.
Moreover, $\lambda$ is open and closed and
$(\lambda(A_i))_{i \in I}$ is a closed, locally finite cover of $KQ(X)$.
Since $\lambda(A_i) = KQ(A_i)$, by the above $\lambda(A_i)$ is $T_2$ for each $i \in I$. Hence, by step 1, $KQ(X)$ is $T_2$ and therefore $X$ is $R_1$.
NOTE: In step 2, the "two element version" is a little bit easier to prove, since one need not show that
$(\lambda(A_i))_{i \in I}$ is locally finite.
