Proof that path-connected set is connected

Question

Rate my proof: Suppose the set $Y$ is path-connected but not connected. Then $Y$ admits a decomposition into disjoint open sets $A, B$. Consider $x \in A, y \in B$, then by path-connectedness there exists a continuous $f:[0,1] \rightarrow Y$ with $f(0)=x, f(1)=y$. We know $[0,1]$ is connected in $\mathbb{R}$, so its image $C=f([0,1])$ is connected in $Y$, therefore $C \subseteq A$ or $C \subseteq B$. Assume WLOG that $C \subseteq A$, then we have $y \in C \subseteq A$, so $y \in A\cap B$, contradicting that $A, B $ are disjoint, so $Y$ is connected. Is this correct?

Answer 1

Yes, the essence is indeed that continuous images of $[0,1]$ (paths) are connected.

We could also choose a path $P_x$ for any $x \in X$ from $p \in X$ fixed and use that $X= \bigcup_{x \in X} P_x$ is a union of all intersecting connected sets e.g.