That's a good question, but you're not correct.
The reason is that if $f\colon\varnothing\to S$, then $f=\varnothing$. Therefore $\breve f=\varnothing$. Therefore $f=\breve f$ and both satisfy the condition for being a function.
It is true, however, that $\breve f$ is not a function whose domain is $S$ (unless $S$ is empty). And note that we shouldn't require it is a function from $S$, because then the function $f(n)=n+1$ as a function from $\Bbb N$ to itself is not injective anymore, which is plain preposterous.
Also, there is but one unique empty function. The empty set.
First note that $\mathbb{W}$ is just shifted $\mathbb{V}$ so regarding the question, there is no difference in those four cases.
1) There exist an injective mapping from $\mathbb{V} × \mathbb{V}$ into $\mathbb{R}$. Since $\mathbb{V} ⊆ \mathbb{R}$, we have a nice injective mapping, even an embedding $\mathbb{V} × \mathbb{V} \to \mathbb{R} × \mathbb{R}$. And there is a theorem stating that for every infinite set $A$, there is a bijection between $A$ and $A × A$. One can imagine the base case of the theorem – enumerating pairs of natural numbers (starting with the origin and concatenating the finite diagonal lines in the first quadrant). So by composing our embedding with a bijection between $\mathbb{R} × \mathbb{R}$ and $\mathbb{R}$, we get desired injective mapping.
2) As said in comments, there is no continuous injective function $\mathbb{V} × \mathbb{V} \to \mathbb{R}$, because the continuous injectivity could then be promoted to embedding on any compact subspace, so we would get a two-dimensional compact subspace of one-dimensional real-line, which is not possible.
3) One can construct an explicit injective function $\mathbb{V} × \mathbb{V} \to \mathbb{R}$. Real numbers are almost the same thing as infinite sequences of ones and zeroes. Informally, you can imagine taking a decimal expansion of the real, but using binary digits instead, the only problem is that there are two different expansions of each rational number – the one ending with $1000…$ and the one ending with $0111…$. Formally, there is a quotient function $2^ω \to [0, 1]$. $2^ω$ is a set of all functions from $ω$ to $2 = \{0, 1\}$, i.e. the set of all infinite sequences of zeroes and ones. It has also a structure of a topological space – taking the product topology of infinite product of $2$-point discrete spaces. This topological space is called Cantor space and is homeomorphic to Cantor set in reals (http://en.wikipedia.org/wiki/Cantor_set). The quotiet just glues the endpoints of the gaps in the Cantor set (i.e. the two equivalent binary expansions of rational numbers) together. So any irrational number has one preimage and any rational number has two preimages. By choosing one preimage for each irrational number, you get an injective function $[0, 1] \to 2^ω$ which is not continuous, but is kind of nice, as created from that quotient mapping. So now we have $[0, 1] × [0, 1] \to 2^ω × 2^ω \to 2^ω \to [0, 1] ⊆ \mathbb{R}$, since $2^ω × 2^ω$ is homeomorphic to $2^ω$ – from a pair of sequences you form one sequence by intertwining the sequences – you realize first sequence on odd indices and the second on even indices.
Best Answer
Into is not a synonym for "injective". There is, however, another way of referring to an injective function: such a function is sometimes said to be "one-to-one function", which is not to be mistaken with a "one-to-one correspondence"/bijective function.
Even though we do refer to a surjective function as being "onto", it does not follow that an injective function is therefore "into."