Yes, $X_\alpha$ can be nonzero for uncountably many $\alpha$. Of course, each individual vector $x \in X$ is zero except for countably many coodinates, in order for $\sum_{\alpha\in\Lambda}\|x_\alpha\|^p<\infty$.

Yes, $X_\alpha$ can be nonzero for uncountably many $\alpha$. Of course, each individual vector $x \in X$ is zero except for countably many coodinates, in order for $\sum_{\alpha\in\Lambda}\|x_\alpha\|^p<\infty$.

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Suppose (for purposes of contradiction) $x \in X$ is such that there are uncontably many $\alpha$ with $\|x_\alpha\| \ne 0$. Then for some $\varepsilon > 0$ there are infinitely many $\alpha$ with $\|x_\alpha\| > \varepsilon$. [Reason: in fact, $\varepsilon$ may be chosen rational, and use the fact that a countable union of finite sets is countable.]
Let $\alpha_1, \alpha_2,\dots$ be such that $\|x_{\alpha_k}\| > \varepsilon$. Then $$ \sum_{\alpha\in \Lambda}\|x_\alpha\|^p \ge \sum_{k=1}^\infty \|x_{\alpha_k}\|^p > \sum_{k=1}^\infty \varepsilon = +\infty . $$ Thus $x \notin X$.