The question is taken from the notes by Joseph L.Taylor http://people.math.sfu.ca/~kya17/seminars/Taylor_Notes_On_Several_Complex_Variables.pdf problem 5.2. Since $D$ is already given to be closed, then $U - D$ is open in $U$ and it suffices to just show that $U - D$ is closed in $U$ as well.
I tried to use the fact from topology that a set is closed iff it contains all its limit points. Let $(\lambda_n)$ denote a sequence in $U - D$ converging to say $\lambda \in U$. We show that $\lambda \notin D$.
By the definition of a subvariety, we have for each $n$, a neighbourhood $W_{\lambda_n}$ of $\lambda_n$ such that $D\cap W_{\lambda_n}=V(S_n)$ for some subset $S_n \subset \mathcal{H}(W_{\lambda_n})$. Unfortunately I am stuck here as I have very little clue on how to use the fact that $\lambda_n \notin D$ to my advantage.
Any help given would be greatly appreciated.
First suppose that $U$ is convex. Take two points $p,q \in U\backslash V$. Draw a (complex) line $L$ joining $p$ and $q$ then the functions defining $V$ restrict to holomorphic functions on $L$ and their common zero set is $V\cap L$. Holomorphy implies that $V\cap L$ is discrete, otherwise $L\subset V$. From $U$ convex we get that $L\cap U$ is connected.
Naturally, the complement of a discrete set in a connected open subset of $\mathbb{C}$ is connected, hence path-connected and you can draw a contiuous path between $p$ and $q$ inside $L\cap U$.
For the general case you can draw a "complex polygonal" and do the same in each edge.