Proving a function is constant in $\mathbb{R}$ if $f(x)=f(2x)$ and f is continuous at zero $f$ is defined as a function on $\mathbb{R}$ which is continuous in 0 and $\forall x \in \mathbb{R}:f(x) = f(2x)$. I do have to proof that $f$ is constant in $\mathbb{R}$.
I figured that this translates to:
$\forall x:f(x) = f(2x) => \forall x \forall y : f(x) = f(y)$
I now want to proof this by contradiction (A = $\neg \neg A$)
$\neg(\forall x:f(x) = f(2x) => \forall x \forall y : f(x) = f(y))$
$\neg (\neg \forall x:f(x) = f(2x) \vee \forall x \forall y : f(x) = f(y)) $
$\forall x : f(x) = f(2x) \wedge \exists x \exists y : f(x) \neq f(y)$
But this does not seem to be sufficient as there could still be a function fulfilling those properties (even though I cannot name one). What am I missing, as I know that such a function cannot exist, but I cannot find any proof within those terms.

Consider any $x \in \mathbb{R}$. Consider the sequence $x_n = \dfrac{x}{2^n}$.
Clearly, we have $\displaystyle \lim_{n \rightarrow \infty} x_n = 0$.
Now, since we have $f(a) = f(2a)$, $\forall a \in \mathbb{R}$, using this, it is easy to prove by induction that $$f(x) = f\left(\dfrac{x}{2^n} \right)$$ $\forall x \in \mathbb{R}$ and $\forall n \in \mathbb{N}$
Now recall that every continuous function is sequentially continuous i.e. $$\displaystyle \lim_{n \rightarrow \infty} f(x_n) = f \left(\lim_{n \rightarrow \infty} x_n \right) $$
Using the above arguments, we get that $\forall x \in \mathbb{R}$, $$f(x) = \displaystyle \lim_{n \rightarrow \infty} f(x) = \displaystyle \lim_{n \rightarrow \infty} f\left(\dfrac{x}{2^n} \right) = f\left(\displaystyle \lim_{n \rightarrow \infty} \dfrac{x}{2^n} \right) = f(0)$$
Hence, $f(x) = f(0)$, $\forall x \in \mathbb{R}$. Hence, $f(x)$ is a constant.
