Lower-order term constraints and wrong guess
First of all, the way you state the induction hypothesis is a bit strange, and it definitely doesn't prove that $T(n) = O(n)$. Assuming that the induction is on $n$, what your induction hypothesis states (more or less) is that for every $n$ there exists $C$ such that $T(k) \leq Ck$ for all $k \leq n$. The function $T(n) = n^2$ satisfies this, with $C = n$.

What you really want to do is to fix the parameters $b,c,n_0$ in advance. In other words, you want to prove
$$
 \exists b,c,n_0 \, \forall n \, \forall n_0 \leq k \leq n\colon T(k) \leq ck - bn.
$$
You first instantiate $b,c,n_0$, and then you prove only the following part by induction:
$$
\forall n \, \forall n_0 \leq k \leq n\colon T(k) \leq ck - bn.
$$
At each step of the induction, you are proving
$$
\forall n_0 \leq k \leq n\colon T(k) \leq ck - bn
$$
for the current value of $n$.

Now for your mistake. Your induction hypothesis states that for all $k \leq n$,
$$ T(k) \leq ck - bn. $$
However, in the inductive step, you only prove
$$ T(k) \leq ck. $$
Also, you only prove it for $k = n$, despite the form of your actual induction hypothesis.

Indeed, it is very strange that your inductive hypothesis gives an upper bound on $T(k)$ which depends on $n$. This is quite unorthodox. Usually the upper bound only depends on $k$. Moreover, you are not really subtracting a lower-order term: $n \neq o(k)$!

Here is what an application of subtracting a lower-order term usually looks like. Suppose that we want to prove that $T(n) = O(n)$. Then you find $b,c,n_0$ and prove the following by induction:

$$ \forall n \geq n_0\colon T(n) \leq bn - c\log n.$$

Note that there is no $k$ here, and that $\log n = o(n)$ is indeed of lower order.