You want to show that there is some $c>0$ such that for all $n$ large enough $T(n) \geqslant cn\lg n$.
I present here a hopefully intuitive inductive approach.
We will first check how the inductive step goes.
Suppose that for some $k\geqslant 1$ we have $T(k)\geqslant ck\lg k$.
Can we show that this relation also holds for some other (greater) values $n$?
If we follow the recursive definition of $T$, we'll see that $T(2k)$ and $T(2k+1)$ can both be expressed in terms of $T(k)$.
First, consider the case $n = 2k$.
Then $T(n) = 2T(k) + 2k$ and we'll have $T(n) \geqslant c(2k)\lg(k) + 2k$.
It's easy to check that $c\leqslant 1$ implies $c(2k)\lg(k)+2k \geqslant c(2k)\lg(2k)$, regardless of the value of $k$.
In this case, $T(n)\geqslant cn\lg n$ as desired.
Now, consider the case $n = 2k+1$.
Then $T(n) = 2T(k) + 2k + 1$ and we'll have $T(n)\geqslant c(2k)\lg(k)+(2k+1)$.
Under what conditions on $c$ do we have
$$c(2k)\lg(k)+(2k+1) \geqslant c(2k+1)\lg(2k+1)\,\,?$$
That is equivalent to
$$c\leqslant \frac{2k+1}{(2k+1)\lg(2k+1)-2k\lg(k)}.$$
You only really need this to be $\geqslant \epsilon>0$ as $k\to\infty$, and it's easy to show that.
For instance, you could show that the RHS converges to $1$ as $k\to\infty$.
That said, you could also show that the minimum is attained at $k=1$ $($with value $\log 2/\log 3\approx 0.63 < 1)$, and this will save us some work in the end with base cases by allowing us to consider base cases right from the get go ($k=1$).
We hence conclude that $c\leqslant \log 2/\log 3$ implies $T(2k+1) \geqslant c(2k+1)\lg(2k+1)$, or $T(n)\geqslant cn\lg n$, once again regardless of the value of $k$.
Combining both cases, we conclude that for $c\leqslant \log 2/\log 3$, $T(k)\geqslant ck\lg k$ implies that both $T(2k)\geqslant c(2k)\lg(2k)$ and $T(2k + 1)\geqslant c(2k + 1)\lg(2k + 1)$.
This will be our inductive step.
To complete the induction, we must now handle the base cases.
Our inductive step shows that the relation can be propagated forward from a base case, but not in the traditional way (from $n$ to $n+1$).
Rather, our inductive step is a sort of doubling ($n$ to $2n$ and $2n+1$), which in general leaves gaps.
Indeed, for the sake of example, suppose we took a single base case of $k = 7$.
Our inductive step would propagate the relation to $n\in\{14, 15\}$, and from these we'd get $n\in\{28,29,30,31\}$, and so on and so forth.
It's easy to see that we'd miss many values.
We need a modified induction for our modified inductive step.
Lemma (Modified induction):
Let $P(n)$ be some proposition about $n$.
Suppose that if $P(n)$ is true, then $P(2n)$ and $P(2n+1)$ are also true.
Additionally, suppose that there is some $j\geqslant 0$, such that $P(n)$ is true for all $n$ with $2^j\leqslant n <2^{j+1}$.
Then $P(n)$ is true for all $n\geqslant 2^j$.
Once you've proved the lemma, we can finish the exercise.
If we take $j=0$ in the lemma, we'd need only show that $P(1)$ is true.
And indeed, it is.
We can calculate the first few values of $T$ by hand: $T(0) = 0$ and $T(1) = 1$.
The proposition $P(1)$ is then $T(1) = 1 \geqslant c\lg 1 = 0$, which is obviously true and concludes the induction.
This approach also helps show where tightness was lacking in the answers provided. Looser values of $c$ (meaning values closer to $1$) will require larger values of $k$ in order for the inductive step $k\mapsto 2k+1$ to work.
And, indeed, for $c\geqslant 1$ (including $c=1$) the inductive step does not hold for any $k$.
