 Solution

Removing $1/2$ from the closed interval $I=[0,1]$ leaves the separation of a topological spaceThus $1/2$ is a cut point of $I$.

In contrast, removing any point $p$ from the square $Q=[0,1{]}^2$ leaves a path-connected space. To prove this, take distinct $u,v\in Q\setminus \{p\}$. Choose an interior point $w$ outside the two lines through $(u,p)$ and $(v,p)$. Such a choice exists: a small circle in the interior meets each of those lines in at most two points. Since $Q$ is a convex set, the line segments $[u,w]$ and $[w,v]$ stay in $Q$; our choice of $w$ ensures that both avoid $p$. Their concatenation is a continuous path from $u$ to $v$ in $Q\setminus \{p\}$.

The product topology on $I\times I$ agrees with its usual subspace topology in the Euclidean plane. If $h:Q\to I$ were a homeomorphism, its restriction would give a homeomorphismThe domain is connected, since path connectedness implies connectedness, while the target is not connected. This contradicts the preservation of connectedness by a homeomorphism.