Your original statement is not true (please read the whole answer). But the following statement is correct:
$$\left \lfloor {\log_2 n} \right \rfloor = \left \lfloor {\log_2 \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor + 1\tag{1}$$
Proof:
Every number $n$ can be placed between two cosecutive powers of 2. In other words, there exists $k$ such that:
$$2^k\le n\le2^{k+1}-1\tag{1}$$
Obviously:
$$k\le\log_2n<k+1$$
$$k=\lfloor\log_2n\rfloor\tag{2}$$
On the other side from (1):
$$\frac{2^k-1}2 \le \frac{n-1}{2} \le \frac{2^{k+1}-2}2$$
$$2^{k-1}-\frac12 \le \frac{n-1}{2} \le 2^k-1$$
$$\lceil 2^{k-1}-\frac12\rceil \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$$2^{k-1} \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$${k-1} \le \log_2\lceil\frac{n-1}{2}\rceil \le \log_2 (2^k-1)<k$$
$${k-1} =\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor$$
$$k=\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1\tag{3}$$
By comparing (2) and (3) you get:
$$\lfloor\log_2n\rfloor = \lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1$$
...which completes the proof.
You can easily prove that the original statement is not true. You are basically saying that the function:
$$f(n)=\left \lfloor {\log n} \right \rfloor - \left \lfloor {\log \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor - 1$$
...is equal to zero for all values of $n$.
This is not true if "$\log$" stands for logartihm with base 10:
This is also not true if "$\log$" stans for natural logarithm "$\ln$":
If you don't trust these plots, calculate the value for $n=45$ and in both cases the result is -1, not 0.
To prove : for $A,B \in \Bbb{R} $
$\lfloor(A + B)\rfloor = \lfloor A\rfloor
+ \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
By definition:
$A = \lfloor A\rfloor + \{A\}.$
$B = \lfloor B\rfloor + \{B\}.$
Therefore,
$$(A + B) = \lfloor A\rfloor + \lfloor B\rfloor
+ \{A\} + \{B\}. \tag1$$
Also, since $0 \leq \{A\},\{B\} < 1$
$0 \leq \{A\} + \{B\} < 2.$
Also, if $P \in \Bbb{R}$ and $z \in \Bbb{Z}$ such that
$0 \leq (P - z) < 1$ then $\lfloor P\rfloor = z.$
Consider two cases separately.
$\underline{\text{Case 1:}~~0 \leq \{A\} + \{B\} < 1}$
Then
$\lfloor \{A\} + \{B\}\rfloor = 0$
Using equation (1) above,
$0 \leq (A + B) - (\lfloor A\rfloor + \lfloor B\rfloor) < 1. $
Therefore,
$\lfloor (A + B)\rfloor = \lfloor A\rfloor + \lfloor B\rfloor =
\lfloor A\rfloor + \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
Therefore, the assertion holds in Case 1.
$\underline{\text{Case 2:}~~1 \leq \{A\} + \{B\} < 2}$
Then
$\lfloor \{A\} + \{B\}\rfloor = 1$
Using equation (1) above,
$0 \leq (A + B) - [(\lfloor A\rfloor + \lfloor B\rfloor) + 1]< 1. $
Therefore,
$\lfloor (A + B)\rfloor = \lfloor A\rfloor + \lfloor B\rfloor + 1 =
\lfloor A\rfloor + \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
Therefore, the assertion holds in Case 2.
Best Answer
Your $\delta$ is not positive if $x$ is an integer (as then $x = \lfloor x \rfloor = \lceil x \rceil$). It is ok to define $\delta $ in terms of $f$, but you have to check that $\delta > 0$.
Note that your proof works for $x \in \mathbb R - \mathbb Z$, so it proves that $\lfloor.\rfloor\colon \mathbb R -\mathbb Z \to \mathbb Z$ is continuous.