Projection of a Jordan curve is injective I'm think the following is true and am hoping someone might be able to provide a reference or counterexample (or a helpful idea):


Let $f:[0,1] \to \mathbb{C}^n$ be a $C^\infty$ a Jordan curve (that is, $f$ is injective on $[0,1)$ and $f(0)=f(1)$.) There exists an $a \in \mathbb{C}^n$ s.t. $P \circ f|_{[0,1)}$ is injective where $P: \mathbb{C}^n \to \{at : t \in \mathbb{C} \}$ is the projection map.


 A: Here's a counterexample.
Let's start with $\mathbb{C}P^{n-1}$, a compact manifold of real dimension $2n-2$ which is the space of $1$-dimensional subspaces of $\mathbb{C}^n$. 
For each $1$-dimensional subspace $S \in \mathbb{C}P^{n-1}$, choose two line segments $\alpha_S=[a_S,b_S]$, $\gamma_S=[c_S,d_S] \subset \mathbb{C}^n$ such that the projection of $\alpha_S,\gamma_S$ to $S$ is a cross. 
We may choose an open neighborhood $U_S \subset \mathbb{C}P^{n-1}$ of $S$, and open neighborhoods $A_S,B_S,C_S,D_S \subset \mathbb{C}P^{n-1}$ of $a_S,b_S,c_S,d_S$ respectively, such that for all $a \in A_S$, $b \in B_S$, $c \in C_S$, $d \in D_S$ and for all $T \in U_S$ the projection of the segments $[a,b]$, $[c,d]$ to $T$ is a cross.  
The sets $U_S$ form an open covering of the compact space $\mathbb{C}P^{n-1}$, and so there exist a finite subset $S_1,...,S_K \in \mathbb{C}P^{n-1}$ such that $U_{S_1},...,U_{S_K}$ cover $\mathbb{C}P^{n-1}$. Consider the points $a_k,b_k,c_k,d_k$ with $a_k = a_{S_k}$ etc. By a small perturbation we may choose $a'_k$, $b'_k$, $c'_k$, $d'_k$ with $a'_k \in A_{S_k}$ etc., so that the collection of segments $2K$ 
$$[a'_1,b'_1],[c'_1,d'_1],...,[a'_K,b'_K],[c'_K,d'_K]
$$
is pairwise disjoint. By construction, the projection of $[a'_k,b'_k]$, $[c'_k,d'_k]$ to each plane in $U_{S_k}$ is a cross. We can then easily connect up these segments to form a simple closed curve, and by construction the projection of this curve to each 1-dimensional subspace contains a cross and is therefore not simple.
