[proofplan] We verify directly that the piecewise-defined concatenation is a continuous map from $[0,1]$ to $X$ with the correct endpoints. The endpoint calculation uses the endpoint data for $\gamma$ and $\eta$. For continuity, we check that the two formulas are continuous on the closed subintervals $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$, agree at the common endpoint, and then prove the required closed-set pasting argument on these two closed pieces. [/proofplan] [step:Define the concatenated map and check that the two formulas agree] Let $A \subset [0,1]$ be the closed subinterval \begin{align*} A=[0,\tfrac{1}{2}], \end{align*} and let $B \subset [0,1]$ be the closed subinterval \begin{align*} B=[\tfrac{1}{2},1]. \end{align*} Define the map \begin{align*} h : [0,1] \to X \end{align*} by the following two prescriptions: for $t \in A$, \begin{align*} h(t)=\gamma(2t), \end{align*} and for $t \in B$, \begin{align*} h(t)=\eta(2t-1). \end{align*} This definition is consistent on $A \cap B=\{\frac{1}{2}\}$. Indeed, \begin{align*} \gamma(2 \cdot \tfrac{1}{2})=\gamma(1)=y \end{align*} because $\gamma$ is a path from $x$ to $y$, while \begin{align*} \eta(2 \cdot \tfrac{1}{2}-1)=\eta(0)=y \end{align*} because $\eta$ is a path from $y$ to $z$. Hence both formulas assign the same value at \begin{align*} t=\tfrac{1}{2}, \end{align*} so $h=\gamma * \eta$ is a well-defined map from $[0,1]$ to $X$. [guided] The only possible ambiguity in the definition of $h$ occurs at the point where the two subintervals overlap. We have \begin{align*} A \cap B=\{\tfrac{1}{2}\}. \end{align*} At this point, the first formula gives \begin{align*} \gamma(2 \cdot \tfrac{1}{2})=\gamma(1). \end{align*} Since $\gamma$ is a path from $x$ to $y$, its terminal value is $\gamma(1)=y$. The second formula gives \begin{align*} \eta(2 \cdot \tfrac{1}{2}-1)=\eta(0). \end{align*} Since $\eta$ is a path from $y$ to $z$, its initial value is $\eta(0)=y$. Thus both formulas give the same value $y$ at \begin{align*} t=\tfrac{1}{2}. \end{align*} This is the compatibility condition that makes the piecewise definition a genuine function rather than two conflicting prescriptions. [/guided] [/step] [step:Verify continuity on each closed subinterval] Define the affine maps \begin{align*} \alpha : A \to [0,1], \quad t \mapsto 2t \end{align*} and \begin{align*} \beta : B \to [0,1], \quad t \mapsto 2t-1. \end{align*} Both $\alpha$ and $\beta$ are continuous because they are restrictions of continuous affine maps $\mathbb{R} \to \mathbb{R}$ to subspaces. Since $\gamma : [0,1] \to X$ and $\eta : [0,1] \to X$ are paths, they are continuous. Therefore the compositions \begin{align*} \gamma \circ \alpha : A \to X \end{align*} and \begin{align*} \eta \circ \beta : B \to X \end{align*} are continuous. By the definition of $h$, the restriction $h|_A$ equals $\gamma \circ \alpha$, and the restriction $h|_B$ equals $\eta \circ \beta$. [/step] [step:Paste the two continuous pieces to obtain continuity on $[0,1]$] We prove the needed pasting argument directly. Let $F \subset X$ be closed with respect to $\tau$. Since $h|_A : A \to X$ is continuous, the set \begin{align*} (h|_A)^{-1}(F) \subset A \end{align*} is closed in the subspace $A$. Since $A$ is closed in $[0,1]$, this set is also closed in $[0,1]$. Likewise, since $h|_B : B \to X$ is continuous and $B$ is closed in $[0,1]$, the set \begin{align*} (h|_B)^{-1}(F) \subset B \end{align*} is closed in $[0,1]$. Now the preimage of $F$ under $h$ decomposes as \begin{align*} h^{-1}(F)=(h|_A)^{-1}(F)\cup(h|_B)^{-1}(F). \end{align*} This is a finite union of closed subsets of $[0,1]$, hence is closed in $[0,1]$. Since the preimage under $h$ of every closed subset of $X$ is closed in $[0,1]$, the map $h:[0,1]\to X$ is continuous. [/step] [step:Compute the endpoints of the concatenated path] At the initial endpoint, \begin{align*} (\gamma * \eta)(0)=h(0)=\gamma(2 \cdot 0)=\gamma(0)=x. \end{align*} At the terminal endpoint, \begin{align*} (\gamma * \eta)(1)=h(1)=\eta(2 \cdot 1-1)=\eta(1)=z. \end{align*} We have shown that $\gamma * \eta=h$ is continuous and has initial point $x$ and terminal point $z$. Therefore $\gamma * \eta$ is a path from $x$ to $z$. [/step]