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Let $G=(V,E)$ be an infinite, vertex-transitive, connected graph of finite degree. Let $0$ be a vertex. Let $d(x,y)$ be the graph distance between two vertices $x$ and $y$, i.e., the length of the shortest path that connects $x$ to $y$. Is it true that for any $x \in V$ there exists $ y \in V$ such that $$ \{y, x\} \in E, $$ and $$ d(y,0) = d(x,0) + 1? $$

It seems to me that it is true in wide generality under the assumptions above (think of $\mathbb{Z}^d$ or trees), but I cannot prove it. Is it maybe necessary to assume a Cayley structure to make it true?

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This seems false for graph with a leaf $x$.Take $o$ to be any other vertex. Then, by definition there is a unique vertex $y$ with $\{x,y\} \in E$ and for such $x$, $d(y,0) = d(x,0) - 1$.

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    What is precisely a graph with a leaf $x$? I asked for infinite, vertex-transitive graphs. – 2017-02-16
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    Perhaps I should have added "connected". I add it in the post. – 2017-02-16
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    I am sorry, I did assume that "vertex-transitive" meant "connected". Do you mean something else ? – 2017-02-16
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    @QuantumLogarithm : a leaf is a vertex of degree $1$. – 2017-02-17