Worst case runtime for binary search
Your algorithm makes $\lfloor \log n  \rfloor+ 1$ comparisons for an array of length $n$, which is at most $2 \log n$, which is $O(\log n)$.  Hence, $\lfloor \log n  \rfloor+ 1 = O(\log n)$.  

In more detail, recall that the big-O notation subsumes constant factors. A function $f(n)$ is said to be $O(\log n)$ if there exist positive constants $c$ and $n_0$ such that $f(n) \le c \log n$ for all $n \ge n_0$.  In our case, we want to prove $\lfloor \log n \rfloor +1 = O(\log n)$.  To prove the existence of positive constants $c$ and $n_0$ satisfying the desired condition, observe that $\lfloor \log n  \rfloor+ 1 \le \log n + \log n = 2 \log n$, for all $n \ge 2$.  So we can take $c = 2$ and $n_0 = 2$.