$X$ is connected if and only if $\beta(X)$ is connected

1.1. \(X\) is embedded as a dense subspace in its Stone-Čech compactification \(\beta(X)\), meaning \(\overline{X} = \beta(X)\).

1.2. If a subspace \(X\) is connected, then its topological closure \(\overline{X}\) in any ambient space is connected.

1.3. Since \(X\) is connected, \(\beta(X) = \overline{X}\) is connected.

2.1. Suppose for contradiction that \(X\) is disconnected, so there exists a separation \(X = A \cup B\), where \(A\) and \(B\) are non-empty, disjoint open (and closed) subsets of \(X\).

2.2. Define \(f: X \to \{0, 1\} \subset [0, 1]\) by:\[f(x) = \begin{cases} 0 & \text{if } x \in A, \\ 1 & \text{if } x \in B. \end{cases}\]

2.3. Since \(A\) and \(B\) are clopen in \(X\), \(f\) is continuous.

2.4. Because \([0, 1]\) is compact Hausdorff, the universal mapping property of the Stone-Čech compactification gives a unique continuous extension:\[\tilde{f}: \beta(X) \to [0, 1] \quad \text{such that } \tilde{f}|_X = f.\]

2.5. The set \(\{0, 1\}\) is closed in \([0, 1]\), so the preimage \(\tilde{f}^{-1}(\{0, 1\})\) is a closed subset of \(\beta(X)\).

2.6. Since \(X \subseteq \tilde{f}^{-1}(\{0, 1\})\) and \(X\) is dense in \(\beta(X)\), we have:\[\beta(X) = \overline{X} \subseteq \tilde{f}^{-1}(\{0, 1\}).\]

2.7. Thus the continuous image \(\tilde{f}(\beta(X))\) is contained in the discrete two-point space \(\{0, 1\}\).

2.8. Because \(\beta(X)\) is connected and \(\tilde{f}\) is continuous, the image \(\tilde{f}(\beta(X))\) must be a connected subset of \(\{0, 1\}\), hence a single point.

2.9. However, \(A \neq \varnothing \implies 0 \in \tilde{f}(\beta(X))\) and \(B \neq \varnothing \implies 1 \in \tilde{f}(\beta(X))\), so \(\tilde{f}(\beta(X)) = \{0, 1\}\), which is disconnected.

2.10. This contradiction shows that \(X\) must be connected.