We clear up some ambiguities in the post and prove it by strong
induction. We let $T(0)=0$ and $T(1)=1$ and prove that when
$$T(n) = \sum_{k=1}^{\lfloor n/2\rfloor}
(-1)^{k+1} {n-k\choose k} T(n-k)$$
for $n\ge 2$ then $$T(n) = C_{n-1}
= \frac{1}{n} {2n-2\choose n-1}
= {2n-2\choose n-1} - {2n-2\choose n}.$$
In fact the case of a zero argument to $T$ is not reached as for $n\ge
2$ we also have $n-\lfloor n/2\rfloor \ge 1.$ Applying the induction
hypothesis on the RHS we get two pieces, the first is
$$A = \sum_{k=1}^{\lfloor n/2\rfloor}
(-1)^{k+1} {n-k\choose k} {2n-2k-2\choose n-k-1}
\\ = {2n-2\choose n-1} +
\sum_{k=0}^{\lfloor n/2\rfloor}
(-1)^{k+1} {n-k\choose k} {2n-2k-2\choose n-k-1}
$$
and the second
$$B = \sum_{k=1}^{\lfloor n/2\rfloor}
(-1)^{k+1} {n-k\choose k} {2n-2k-2\choose n-k}
\\ = {2n-2\choose n} +
\sum_{k=0}^{\lfloor n/2\rfloor}
(-1)^{k+1} {n-k\choose k} {2n-2k-2\choose n-k}.$$
As we subtract $B$ from $A$ we see that we only need to show that the
contribution from the two sum terms is zero.
For these two pieces we introduce the integral representation
$${n-k\choose k} = {n-k\choose n-2k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2k+1}} (1+z)^{n-k} \; dz.$$
This has the nice property that it vanishes when $k\gt \lfloor
n/2\rfloor$ so we may extend the upper limit of the sum to infinity.
We also introduce for the first sum
$${2n-2k-2\choose n-k-1} =
\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n-k}} (1+w)^{2n-2k-2} \; dw.$$
We thus obtain
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n}} (1+w)^{2n-2}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n}
\sum_{k\ge 0} (-1)^{k+1} \frac{z^{2k} w^k}{(1+z)^k (1+w)^{2k}}
\; dz\; dw
\\ = - \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n}} (1+w)^{2n-2}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n}
\frac{1}{1+z^2 w/(1+z)/(1+w)^2}
\; dz\; dw
\\ = -\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n}} (1+w)^{2n}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n+1}
\frac{1}{(1+z)(1+w)^2+z^2 w}
\; dz\; dw
\\ = -\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} (1+w)^{2n}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n+1}
\frac{1}{z+1+w} \frac{1}{z + (1+w)/w}
\; dz\; dw.$$
We evaluate the inner integral by summing the residues at $z=-(1+w)$
and $z=-(1+w)/w$ and flipping the sign. (We will verify that the
residue at infinity is zero.)
The residue at $z=-(1+w)$ yields
$$-\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} (1+w)^{2n}
\\ \times \frac{(-1)^{n+1}}{(1+w)^{n+1}} (-1)^{n+1} w^{n+1}
\frac{1}{-(1+w)+(1+w)/w} \; dw
\\ = -\frac{1}{2\pi i}
\int_{|w|=\gamma} (1+w)^{n-1} \frac{w}{1-w^2}\; dw.$$
This is zero as the pole at zero has been canceled. Next for the residue
at $z=-(1+w)/w$ we get
$$-\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} (1+w)^{2n}
\\ \times \frac{(-1)^{n+1} w^{n+1}}{(1+w)^{n+1}}
(-1)^{n+1} \frac{1}{w^{n+1}} \frac{1}{-(1+w)/w+1+w} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} (1+w)^{n-1}
\frac{w}{1-w^2} dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n}} (1+w)^{n-2}
\frac{1}{1-w} dw.$$
With $n\ge 2$ we can evaluate this as
$$\sum_{q=0}^{n-1} {n-2\choose q} = 2^{n-2}.$$
To wrap up the residue at infinity of the inner integral is
$$\mathrm{Res}_{z=\infty} \frac{1}{z^{n+1}} (1+z)^{n+1}
\frac{1}{z+1+w} \frac{1}{z + (1+w)/w}
\\ = -\mathrm{Res}_{z=0} \frac{1}{z^2} z^{n+1}
\frac{(1+z)^{n+1}}{z^{n+1}}
\frac{1}{1/z+1+w} \frac{1}{1/z + (1+w)/w}
\\ = -\mathrm{Res}_{z=0} (1+z)^{n+1}
\frac{1}{1 + z (1+w)} \frac{1}{1 + z (1+w)/w} = 0.$$
Collecting everything and flipping the sign we have shown that
$$A = - 2^{n-2}.$$
For piece $B$ we see that it only differs from $A$ in an extra $1/w$
factor on the extractor in $w$ at the front. We thus obtain
$$-\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+2}} (1+w)^{2n}
\\ \times \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n+1}
\frac{1}{z+1+w} \frac{1}{z + (1+w)/w}
\; dz\; dw.$$
The residue at $z=-(1+w)$ vanishes the same because there was an extra
$w$ to spare on the $w/(1-w^2)$ term:
$$-\frac{1}{2\pi i}
\int_{|w|=\gamma} (1+w)^{n-1} \frac{1}{1-w^2}\; dw.$$
For the residue at $z=-(1+w)/w$ we are now extracting from
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} (1+w)^{n-2}
\frac{1}{1-w} dw.$$
to get
$$\sum_{q=0}^{n} {n-2\choose q} = 2^{n-2}$$
as before. The residue at infinity vanished in $z$ and did not reach
the front extractor in $w$, for another contribution of zero. This
means that
$$B = - 2^{n-2}$$
and we may conclude the proof. The fact that the sum term from the
geometric series factored as it did is the remarkable feature of this
problem.
