The following is an argument that is in spirit close to a sketch once posted by boumol. In some ways it does not differ very much from the solution by JDH, but it has a more syntactic feel.
Let $Succ(x,y)$ be a formula that says that $y$ is "the successor" of $x$; for example, it could be an abbreviation for $(x \le y) \land \forall u((x \le u \land u \le y) \longrightarrow (u=x)\lor (u=y)).$
Consider the following axioms, the conjunction of which will be the sentence $\phi$ asked for in the post. The axioms are given semi-formally, but in a way that is easily made fully formal.
There is a smallest element, $\exists u\forall v(u \le v)$.
The smallest element has a successor.
If $x$ is a successor then $x$ has a successor, $\forall x(\exists u Succ(u,x) \longrightarrow \exists v Succ(x,v))$.
If $x$ has a successor and $x$ is not the smallest element, then there is a $w$ such that $x$ is the successor of $w$.
There is an object which is not a successor, and is not the smallest element.
If $x$ is not a successor, and $x$ is not the smallest element, then there is a non-successor $u$ such that $u$ is not the smallest element, and such that $u \le x$, and $u \ne x$.
It is easy to come up with models for the conjunction $\phi$ of these axioms. A simple one is the natural numbers followed by the open interval $(0,1)$, under the obvious order. This model had already been mentioned by boumol.
Now we show that the above set of axioms does not have a model which is a closed set of reals. Suppose to the contrary that the closed set $K$ of reals, with the natural order, is a model. Without loss of generality we may assume that the smallest element of $K$ is $0$.
It is easy to see that any model of the axioms has an initial segment that is order isomorphic to $\mathbb{N}$.
Let $D\subset K$ consist of all elements of $K$ other than $0$ which are non-successors.
Then $D$ is non-empty and bounded below. Let $m=\inf D$. By Axiom 6, $m$ must be a successor. Also, every object smaller than $m$ other than $0$ is a successor, by the definition of $m$.
It is clear that $m$ cannot belong to the initial segment $K_0$ of $K$ which is isomorphic to $\mathbb{N}$. So there are successors less than $m$ which are bigger than any element of $K_0$.
Let $a$ be the supremum of $K_0$. Then $a \in K$, since $K$ is closed. By the definition of $m$, this number $a$ must be a successor. But then the predecessor of $a$ is an upper bound for $K_0$, contradicting the fact that $a$ is the supremum of $K_0$.