there is no convexity assumption needed there.

Nevertheless here is a counterexample of $1$-Lipschitz function on $\mathbb{R}^{N}$ which fails to satisfy concentration inequality on the Hamming cube $\{-1,1\}^{N}$ as $N$ goes to infinity.

Take $N$ to be even. Let
$$
A = \left\{ (x_{1}, \ldots, x_{N}) \in \{0,1\}^{N}\, : x_{1}+\ldots+x_{N}\leq \frac{N}{2}\right\}.
$$
Next, define a function

$$
f(x) = \inf_{y \in A} \| x-y\|_{\mathbb{R}^{N}}
$$
Since $f$ is the distance function to a nonempty subset it follows that $f$ is $1$-Lipschitz (an exercise).

On the other hand notice that $f(x) = \sqrt{\max\{x_{1}+\ldots+x_{N} - \frac{N}{2},0\}}$ on $\{0,1\}^N$. Then as $N$ goes to infinity we have
$$
\begin{aligned}
P(|f & -\mathbb{E}f| > N^{1/4}) \\ & = P\biggl(\biggl|\sqrt{\max\{\frac{(2x_{1}-1)+\ldots+(2x_{N}-1)}{\sqrt{N}},0\}} \\ & \qquad\qquad - \mathbb{E}\sqrt{\max\{\frac{(2x_{1}-1)+\ldots+(2x_{N}-1)}{\sqrt{N}},0\}} \biggr|>\sqrt{2} \biggr) \\ & \to P\left( | \sqrt{\max\{\xi, 0\}} - \mathbb{E} \sqrt{\max\{\xi, 0\}}|>\sqrt{2}\right)>10^{-10},
\end{aligned}
$$
where we used the central limit theorem, and $\xi$ is the standard normal Gaussian $\xi\in N(0,1)$.