Hölder 1/2 condition for curve Does there exist a continuous  map $\gamma:[0,1] \to \mathbb{R}^n$ s.t. $$|\gamma(x)-\gamma(y)| \ge |x-y|^{\frac 1 2}  $$ for all $x,y \in [0,1]$?
 A: Yes, there is such a function. You should be able to construct one explicitly by an appropriate modification of the Koch snowflake construction, although I don't want to work through the details.
In fact, much more is true. Take any metric space $(X,d)$ which satisfies a certain finite-dimensionality condition called "doubling". (This means that, for some constant $N$, every ball of radius $r$ can be covered by at most $N$ balls of radius $r/2$.)
Now take any $0<\alpha <1$. Then there is a constant $C\geq 1$, a positive integer $n$, and a map $f:X\rightarrow \mathbb{R}^n$ which satisfies
$$ C^{-1} d(x,y)^\alpha \leq |f(x) - f(y)| \leq Cd(x,y)^\alpha.$$
This is called Assouad's embedding theorem.
What does this have to do with your question? Well, let $(X,d)$ be $[0,1]$ with the usual metric given by $|\cdot|$. This is a doubling metric space, so by Assouad's theorem (with $\alpha=1/2$) we get a map $f:[0,1]\rightarrow\mathbb{R}^n$ satisfying
$$ C^{-1} |x-y|^{1/2} \leq |f(x) - f(y)| \leq C|x-y|^{1/2}.$$
If you just rescale $f$ by a factor of $C$, you get
$$ |x-y|^{1/2} \leq |f(x) - f(y)| \leq C^2|x-y|^{1/2},$$
which in particular is what you wanted.
A: While Heinonen's exposition of Assouad's theorem is excellent, one should not forget the original 1983 paper by Assouad, Plongements lipschitziens dans $\mathbb R^n$. Proposition 4.4 answers your question in a more general form. I restate it below, changing terminology to make it self-contained:

Let $k\ge 1$ be an integer, and  $p\in (0,1)$. Suppose $n>1/p$ is an integer. Then there exists a map $f:[0,1]^k\to \mathbb R^n$ such that 
  $$c|x-y|^p \le |f(x)-f(y)|\le C|x-y|^p \tag1$$
  for all $x,y\in [0,1]^k$, with some positive constants $c$ and $C$.

You can take $k=1$, $p=1/2$, and $n=3$. Then multiply $f$ by a constant to get $c=1$.  
As GCD hinted, the construction by Assouad is an   $n$-dimensional form of von Koch snowflake.
It is known that $f$ as in (1) does not exist for $n=1/p$ (in particular, when $n=2$ and $p=1/2$). Indeed, (1) implies that the range of $f$ must be a porous set and therefore its $n$-dimensional measure must be zero; when $n=1/p$ the latter is incompatible with the lower bound in (1). See Porous sets and quasisymmetric maps by Väisälä. 
The above paragraph leaves open the question of whether you can take $n=2$ if you only want the lower bound in (1) (plus continuity of $f$). I do not know the answer, but suspect it is negative. 
One more reference: Geometric embeddings of metric spaces by Heinonen, specifically Chapter 3.
