Let $G$ be an $n$-vertex graph with $n\geq 3$. If for every (ordered) triple of vertices, $(x,y,z)$, there exists an $x,z$-path through $y$, then $G$ is $2$-connected.
Proof: Assume that for every (ordered) triple of vertices, $(x,y,z)$, there exists an $x,z$-path through $y$. Hence, the order $(y,x,z)$ ensures that there exists a $y,z$-path containing $x$. This implies that there exists an $x,z$-path that does not contain $y$, as paths do not duplicate vertices. Since $y$ is an arbitrary vertex that is contained in an $x,z$-path, this is true for all vertices $y\in V(G)$ such that $y\neq x, y\neq z$.
Thus, by appending the $x,z$-path through $y$ to the $x,z$-path which does not contain $y$ yields a cycle through $x$ and $z$. We conclude that for any $x,z\in V(G)$, there exists a cycle containing $x$ and $z$. Therefore, $G$ is 2-connected.
I'm not sure if I am leaving a gap when I make the statement 'since y is an arbitrary vertex that is contained in an $x,z$-path, this is true for all $y\in V(G):y\neq x, y\neq z$.'
I realize that you were looking for proof verification and that seems tough [see my comment above]. There is however, an easier way to prove this, and it may be the desired proof:
Suppose $G$ is not 2-connected. Then $G$ has a cut vertex $x$; i.e., there is a vertex $x$ such that $G \setminus \{x\}$ has at least two components. Then let $z \not =x $ be in one component and $y \not =x$ be in another component. Then there is no path $P_{xzy}$ from $x$ to $z$ that contains $y$. [Indeed, if there were such a path $P_{yxz}$ then there are paths $P_{xy}$ and $P_{yz}$ from $x$ to $y$, and $y$ to $z$, that intersect precisely at $y$. But as $x$ is a cut vertex and $y$ and $z$ are in different components of $G \setminus \{x\}$ it follws that $P_{yz}$ has to also contain $x \in P_{xy}$ as well.]
Thus, from the above, we note: if $G$ is not 2-connected, then there is an ordered triple $(x,y,z)$ such that $G$ has no path from $x$ to $z$ containing $y$. The result follows.
You could also assume that $G$ has at least 4 vertices and, this time writing as $x'$ the cut vertex, have $x$, $y$, and $z$ be such that $x$ and $y$ are in different components of $G \setminus \{x'\}$, and then $y$ and $z$ in different components of $G \setminus \{x'\}$ [it doesn't matter if $x$ and $z$ are in the same component of $G \setminus \{x'\}$.]