Question clarification. This is an exercise question:
"Let $V$ be an open subset of $\mathbb{R}^3$. Does there exist a continuous injective function $f:V \rightarrow \mathbb{R}$?"
What exactly does it mean?
Is the question: "Given an open subset of $\mathbb{R}^3$, does there always exist a continuous injective function $f:V \rightarrow \mathbb{R}$?"?
$$
\forall V\subseteq \mathbb{R}^3:\ V \text{ is open } \Rightarrow \exists f:\ V \rightarrow \mathbb{R} \text{ continuous and injective}
$$
Or is it maybe:
"Does there exist a continuous injective function $f:V \rightarrow \mathbb{R}$? where $V$ is an open subset of $\mathbb{R}^3$.
$$
\exists f:\ V \rightarrow \mathbb{R} \text{ continuous and injective}, V\subseteq \mathbb{R}^3:\ V \text{ is open }
$$
I honestly do not have a clue.
How I would go about (dis)proving this would very much depend on my understanding.
 A: The statement can be reformulated to

Let $V\subset \mathbb R^3$ be open. Under what circumstances does there exist a continuous injective function $f:V\to\mathbb R$?

If you really want to make it formal, you can write

Find a property $P$ such that
  $$\forall V \subset \mathbb R^3 \text{ open}: P(V) \implies \exists f:V\to \mathbb R\text{ continuous and injective}$$

The answer would then be $P(V) := V = \emptyset$.

If $V$ is the empty set, there is such an $f$ (called the empty function).
It doesn't map anything and since $\mathrm{dom}(f) = \emptyset$, all statements about preimages of $f$ are vacuously true, hence $f$ is injective.
If $V$ isn't empty, it contains some open ball $B_\epsilon(x_0) \subset \mathbb R^3$. Since there is no injective map from $B_1(0)$ to $\mathbb R$ (and any $B_\epsilon(x_0)$), there cannot be an injective map from $V$ to $\mathbb R$.
Thus, the answer to the question is:

If $V\subset \mathbb R^3$ is open, there exists a continuous function $f:V\to\mathbb R$ if and only if $V=\emptyset$.

A: Use the following fact:

There is no locally injective continuous function $f:[0,1] \to \Bbb{R}$ such that $f(0)=f(1)$.

Now, since $V$ is open, you can build a closed curve (for example, the equator of a sphere) inside some ball contained in $V$. This means that you are able to find some $\gamma: [0,1] \to V$ with $\gamma(0)=\gamma(1)$, which parametrizes your curve. You can suppose WLOG that $\gamma|_{[0,1)}$ is injective.
If $f: V \to \Bbb{R}$ is injective and continuous, then $f \circ \gamma$ is a locally injective continuous function and $f \circ \gamma(0) = f \circ \gamma(1)$: a contradiction.
