Using proof by induction to solve recurrence relation? 
$$T(n) = \begin{cases} 1 & \text{ if } n = 1,\\\\
    T(n/2)+d & \text{ if } n > 1 \\\\
    \end{cases} 
    $$

is $T(n) \leq c\log_2 n$
I'm still new to recurrence relations, so any help would be great! Thanks in advance!

Basis:
We check if the relation $T(n) \leq c \log_2 n \ \ \ (*)$ stands for $n=1$.
From the recursive relation we have that $T(1)=1 \geq c \log_2 1=0$
So the relation $(*)$ does not stand for $n=1$.
The asympotic means that $T(n) \leq c \log_2 n, \forall n \geq n_0$, where $n_0$ is a constant.
We check if the relation $(*)$ stands for $n=2$.
From the recursive relation we have that $T(2)=T(1)+d=1+d$. Now we check if $\exists c>0$ such that $T(2) \leq c \log_2 2 \Rightarrow 1+d \leq c \Rightarrow c \geq 1+d \ \ \ \checkmark$
We check if the relation $(*)$ stands for $n=3$.
From the recursive relation we have that $T(3)=T(\lfloor \frac{3}{2} \rfloor )+d=T(1)+d=1+d \leq c \log_2 3$. This stands if $c \geq 1+d$.
Hypothesis:
Let $n>2$. We suppose that the relation $(*)$ stands for each $m$ such that $2 \leq m \leq n$, that means that we suppose that $T(m) \leq c \log_2 m, \ \ \ \forall m , 2 \leq m \leq n$, where $c \geq 1+d$ a constant.
Inductive step:
We will show that the assumption stands for $n$, that means that we will show that $T(n) \leq c \log_2 n$.
We have already shown that it stands for $n=3$.
For $n>3$:
From the recursive relation we have that $T(n)=T(\lfloor \frac{n}{2} \rfloor)+d$
Since $n>3 \Rightarrow \lfloor \frac{n}{2} \rfloor \geq 2$.
From the hypothesis, for $m=\lfloor \frac{n}{2} \rfloor$:
$T(\lfloor \frac{n}{2} \rfloor) \leq c \log_2{(\lfloor \frac{n}{2} \rfloor)} \leq c(\log_2 n-\log_2 2)=c(\log_2 n-1)$
So, $T(n) \leq c(\log_2 n-1)+d =c \log_2n -(c-d) \leq c \log_2 n$, if $c-d \geq 0 \Rightarrow  c \geq d $
If we choose $c=1+d$ the assumption is true.
