Is it true that a connected graph has a spanning tree, if the graph has uncountably many vertices?

I figured out the answer and someone asked me to post it as an answer, so I am answering my own question in case this is helpful for anyone else.

Let $G$ be a connected graph and consider the set $S$ of all trees $T \subset G$ ordered by the subgraph relation. We wish to apply Zorn's lemma, so we must check first that every chain $\mathcal{C}$ has an upper bound, i.e. a tree containing every $T \in \mathcal{C}$ as a subgraph. We claim that $T^* = \cup \mathcal{C}$ is just such a tree.

Clearly $T$ a subgraph of $T^*$ for every $T \in \mathcal{C}$ by construction. The main thing is to check that $T^*$ is a tree. Suppose $T^*$ is not a tree. Then either $T^*$ is disconnected or it has a cycle. If $T^*$ is disconnected, then there are two vertices $u,v \in T^*$ which are not connected by a path in $T^*$. But $u \in T_u$ for some $T_u \in \mathcal{C}$ and $v \in T_v$ for some $T_v \in \mathcal{C}$ and $\mathcal{C}$ is a chain, so either $T_u \subset T_v$ or $T_v \subset T_u$ and in either case $u$ is connected to $v$ by a path in the larger tree, and $T^*$ in turn contains this path, a contradiction. The contradiction to $T^*$ containing a cycle is similar --- every edge in the cycle must appear somewhere in the chain, and then one of the supposed trees in the chain is not acyclic.

Every chain has an upper bound, so $S$ has a maximal element, by Zorn's lemma. It is easy to see that it must be a spanning tree.