Does this collection form an algebra over $\mathbb{R}$? Does the collection
$$\mathcal{A}=\{V\cup K\mid V\subset\mathbb{R}\textrm{ is open and } K \subset \mathbb{R} \textrm{ is closed}\}$$
form an algebra over $\mathbb{R}$?

No, it does not.  Note that every open set $U$ is in $\mathcal{A}$ since $U = U \cup \emptyset$ and $\emptyset$ is closed; and similarly, every closed set $K$ is in $\mathcal{A}$.  Now, if $\mathcal{A}$ were an algebra, that would imply every intersection of an open set and a closed set in $\mathbb{R}$ would also be a union of an open set and a closed set.
We now give a counterexample to show this is not true.  Consider the open set $U = (0, \infty)$ and the closed set $K = \{ 0 \} \cup \{ \frac{1}{n} : n \in \mathbb{N}_+ \}$.  Then $U \cap K = \{ \frac{1}{n} : n \in \mathbb{N}_+ \}$.  Suppose we had $U \cap K = V \cup L$ where $V$ is open and $L$ is closed.  Then for each $n$, $\frac{1}{n} \notin V$ since otherwise $U \cap K = V \cup L$ would contain an open neighborhood of $\frac{1}{n}$, which is a contradiction.  Therefore, $V = \emptyset$, and it follows that $L = \{ \frac{1}{n} : n \in \mathbb{N}_+ \}$.  However, then $L$ is not closed, giving the desired contradiction.
