About the norm in $C^{\ell, \alpha}(\overline{\Omega})$.

Question

In my functional analysis course, we defined for $\ell \in \mathbb N$ and $\alpha \in (0, 1]$, \begin{align*} C^{\ell, \alpha}(\Omega) &= \left\{u \in C^\ell(\Omega)~\bigg|~ \max_{|\beta| = \ell}\sup_{x \neq y \in \Omega} \frac{|\partial^\beta u(x) - \partial^\beta u(y)|}{|x - y|^\alpha} < +\infty\right\},\\ C^{\ell, \alpha}(\overline{\Omega}) &= \left\{u ~\bigg|~ \exists U \text{ open }, \overline{\Omega} \subset U, u \in C^{\ell, \alpha}(U)\right\}. \end{align*} For $\Omega$ bounded, my teacher said (an Wikipedia agreed) that these sets are Banach spaces with respect to the norm $$\|u\|_{C^{\ell, \alpha}} = \max_{|\beta| \le \ell}\|\partial^\beta u\|_{L^\infty(\Omega)} + \max_{|\beta| = \ell}\sup_{x \neq y \in \Omega} \frac{|\partial^\beta(x) - \partial^\beta(y)|}{|x - y|^\alpha}.$$ But on $C^{\ell, \alpha}(\overline{\Omega})$, I don't see why $\|\cdot\|_{C^{\ell, \alpha}}$ is even a norm. Indeed, for $\ell = 0, \alpha = 1$ and $\overline{\Omega} = [0, 1]$, we can take $$ u(x) = \begin{cases} 0 & \text{if } x \in (-1, 1],\\ x - 1 & \text{if } x \in (1, 2). \end{cases} $$ If I am not mistaken, $u \in C^{0,1}((-1, 2))$ and $\|u\|_{C^{0, 1}} = 0$ but $u \neq 0$. Perhaps we should consider $\|u\|_{C^{0, 1}(U)}$ but now I'm not sure that everything is well-defined since for one $u$ there could be $\overline{\Omega} \subset U_1 \neq U_2$ open such that $\|u\|_{C^{0, 1}(U_1)} \neq \|u\|_{C^{0, 1}(U_1)}$.