I proved one of Hatcher's propositions on my own and my proof is quite a bit different than his.
The Unique Lifting Property says: Given a covering space $p:\tilde{X} \rightarrow X$ and a map $f: Y \rightarrow X$, if $\tilde{f}_{1}, \tilde{f}_{2}: Y \rightarrow \tilde{X}$ are lifts of $f$ which agree at one point of $Y$ and $Y$ is path connected, then $\tilde{f}_{1}$ and $\tilde{f}_{2}$ agree at all points of $Y$.
My proof is as follows:
Let $A=\{y \in Y : \tilde{f}_{1}(y)=\tilde{f}_{2}(y)\}$ and $B=\{y \in Y : \tilde{f}_{1}(y)\neq \tilde{f}_{2}(y)\}$; then $Y=A \cup B$. Note that $A \neq \emptyset$ by hypothesis. Suppose $y \in A$ and consider $f(y) \in X$. Let $U$ be a neighborhood of $f(y)$ which is evenly covered. Let $\tilde{V}$ be the sheet above $U$ containing $\tilde{f}_{1}(y)=\tilde{f}_{2}(y)$. Let $G=\tilde{f}^{-1}_{1}(\tilde{V}) \cap \tilde{f}^{-1}_{2}(\tilde{V})$; suppose $t \in G$ then $p(\tilde{f}_{1}(t))=p(\tilde{f}_{2}(t))=f(t)$. Since $\tilde{f}_{1}(t),\tilde{f}_{2}(t)\in V$ and $p$ is injective on $V$, we have $\tilde{f}_{1}(t)=\tilde{f}_{2}(t)$. Thus $G$ is an open subset of $A$ containing $y$.
Suppose now that $y \in B$ and let $U$ be a neighborhood of $f(y)$ which is evenly covered. Let $\tilde{V}_{1}$ and $\tilde{V}_{2}$ be the sheets above $U$ containing $\tilde{f}_{1}(y)$ and $\tilde{f}_{2}(y)$, respectively. Note that $\tilde{V}_{1}\neq \tilde{V}_{2}$ as this would imply that $\tilde{f}_{1}(y)=\tilde{f}_{2}(y)$. Let $G=\tilde{f}^{-1}_{1}(\tilde{V}_{1}) \cap \tilde{f}^{-1}_{2}(\tilde{V}_{2})$. If $t \in G$ then $\tilde{f}_{1}(t) \in \tilde{V}_{1}$ and $\tilde{f}_{2}(t) \in \tilde{V}_{2}$ so they are not equal. Thus $G$ is an open subset of $B$ containing $y$. It follows that $B = \emptyset$ or this would be a separation of $Y$, contradicting that $Y$ is connected.
Does this proof seem correct?
Best Answer
[I'm not allowed to comment due to missing reputation, but my post might be considered a comment.]
Your proof seems similar to the one given by Edwin H. Spanier in his book "Algebraic Topology". I'm including a screenshot of his proof for reference.