[proofplan] We prove the two implications directly from the definition of path-connectedness. The forward direction is obtained by applying path-connectedness to the ordered pair $(x_0,x)$. For the reverse direction, paths from the basepoint to two arbitrary points are supplied by hypothesis; reversing the first path and concatenating it with the second gives a path between the arbitrary pair. [/proofplan]

[step:Use path-connectedness to produce paths from the basepoint] Assume that $X$ is path-connected. Let $x \in X$ be arbitrary. By the definition of path-connectedness, there exists a continuous map $\gamma : [0,1] \to X$ such that $\gamma(0)=x_0$ and $\gamma(1)=x$. Hence every point of $X$ can be joined to $x_0$ by a path starting at $x_0$. [/step]

[step:Reverse and concatenate basepoint paths to connect arbitrary points]Assume that for every $x \in X$, there exists a continuous map $\gamma : [0,1] \to X$ such that $\gamma(0)=x_0$ and $\gamma(1)=x$. Let $x,y \in X$ be arbitrary. By the hypothesis applied to $x$, choose a path $\gamma_x : [0,1] \to X$ from $x_0$ to $x$. By the hypothesis applied to $y$, choose a path $\gamma_y : [0,1] \to X$ from $x_0$ to $y$. By [citetheorem:9158], the reversed path $\overline{\gamma_x}: [0,1] \to X$, defined by $\overline{\gamma_x}(t)=\gamma_x(1-t)$, is a path from $x$ to $x_0$. Since $\overline{\gamma_x}$ ends at $x_0$ and $\gamma_y$ begins at $x_0$, [citetheorem:9159] implies that the concatenation $\overline{\gamma_x} * \gamma_y : [0,1] \to X$ is a path from $x$ to $y$.[/step]

[guided]We must prove path-connectedness, so we begin with an arbitrary ordered pair of points $x,y \in X$ and construct a path from $x$ to $y$. The hypothesis gives paths only from the fixed basepoint $x_0$ outward, so it gives a path $\gamma_x : [0,1] \to X$ with $\gamma_x(0)=x_0$ and $\gamma_x(1)=x$, and a path $\gamma_y : [0,1] \to X$ with $\gamma_y(0)=x_0$ and $\gamma_y(1)=y$. The first path has the wrong direction for connecting $x$ to $y$: it goes from $x_0$ to $x$, but we need to start at $x$. We therefore reverse it. By [citetheorem:9158], the map $\overline{\gamma_x}: [0,1] \to X$ defined by $\overline{\gamma_x}(t)=\gamma_x(1-t)$ is continuous, satisfies $\overline{\gamma_x}(0)=x$, and satisfies $\overline{\gamma_x}(1)=x_0$. Thus $\overline{\gamma_x}$ is a path from $x$ to $x_0$. Now the endpoint of $\overline{\gamma_x}$ equals the starting point of $\gamma_y$, namely $x_0$. This is exactly the endpoint-matching condition required for path concatenation. By [citetheorem:9159], the concatenated map $\overline{\gamma_x} * \gamma_y : [0,1] \to X$ is continuous and is a path from $x$ to $y$.[/guided]

[step:Conclude that the space is path-connected] The preceding step shows that for every pair $x,y \in X$, there exists a path in $X$ from $x$ to $y$. Therefore, by the definition of path-connectedness, $X$ is path-connected. Combining this with the first implication proves the equivalence. [/step]