You've omitted a few steps. It looks like you're attempting to prove by induction that $T(n) = O(n)$, and your proof goes:
Suppose $T(k) = O(k)$ for $k<n$. This means $T(k) \le c \, k$ for some $c$. Then $T(n) = 2 T(\lfloor n/2\rfloor) + n \le 2 c \lfloor n/2\rfloor + n\le (c+1) \,n$, so $T(n) = O(n)$.
This proof goes wrong right from the start: “$T(k) = O(k)$ for $k \lt n$” does not make sense. Big oh is an asymptotic notion: $T(k) = O(k)$ means that there is some constant $c$ and a threshold N such that $\forall k \ge N, T(k) \le c \, k$. And again at the end, you can't conclude that “$T(n)=O(n)$”, because that says something about the function $T$ as a whole and you've only proved something about the particular value $T(n)$.
You need to be explicit about what $O$ means. So maybe your proof goes:
Suppose that $T(k) \le c \, k$ for all $k < n$. Then $T(n) = 2 T(\lfloor n/2\rfloor) + n \le 2 c \lfloor n/2\rfloor + n\le (c+1) \,n$.
This does not prove an inductive step: you started from $T(k) \le c \, k$, and you proved that for $k=n$, $T(k) \le (c+1) \, k$. This is a weaker bound. Look at what this means: $T(k) \le c \, k$ means that $c$ is a bound for the rate of growth of $T$. But you have a rate $c$ that grows when $k$ grows. That's not a linear growth!
If you look closely, you'll notice that the rate $c$ grows by $1$ whenever $k$ doubles. So, informally, if $m=2^p k$ then $c_m = c_k + p$; in other words, $c_k = c_0 \log_{2} k$.
This can be made precise. Prove by induction that for $k \ge 1$, $T(k) \le c \log_2(k)$.
The recurrence relation is typical for divide-and-conquer algorithms that split the data in two equal parts in linear time. Such algorithms operate in $\Theta(n\,\mathrm{log}(n))$ time (not $O(n)$).
To see what the expected result is, you can check the recurrence relation against the master theorem. The division is $2T(n/2)$ and the extra work done is $n$; $\log_2(2) = 1$ so this is the second case for which the growth is $\Theta(n\,\log(n))$.