TITLE: Does path-connected imply simple path-connected?
QUESTION [24 upvotes]: Let $X$ be a path-connected topological space, i.e., for any two points $a,b\in X$ there is a continuous map $\gamma\colon[0,1]\to X$ such that $\gamma(0)=a$ and $\gamma(1)=b$. Note that beyond continuity little is required about $\gamma$.

Is it always possible to make $\gamma$ a simple (=injective) curve? (Of course we consider only the case $a\ne b$). 
  If the answer depends on $X$, what are some mild sufficient conditions on $X$?

If $X$ is sufficiently nice, e.g., an open subset of $\Bbb R^n$, the answer is "yes";  more generally, what one would call "locally simplepath-connected" suffices (the set of points reachable from $a\in X$ by a simple curve is both open and closed).
Also, for arbitrary $X$ it is clear that a single instance of self-crossing can be short-cut. But for infinitely many possible short-cuts the situation may become somewhat hairy, I'm afraid.
EDIT: To avoid Mike Miller's counterexample (indiscrete topology on $X$ with $|X|<|\Bbb R|$), let's at least assume that $X$ is Hausdorff. At any rate, $|X|\ge|\Bbb R|$ is a necessary condition - I am still looking for "as mild as possible"  sufficient conditions ...

REPLY [2 votes]: Since the OP asks for conditions, which are "mild as possible," I'll point out that the Hausdorff condition can actually be weakened slightly.
Definition: A space $X$ is $\Delta$-Hausdorff if the image of every path $\alpha:[0,1]\to X$ is closed in $X$.
We have the following.
Theorem: Every path-connected, $\Delta$-Hausdorff space is arcwise connected.
Proof. You can find the proof in this expository note.
At its heart, this proof is basically the one given in the accepted answer, where the idea is to make choices of loops to delete. Although the converse of the theorem is certainly false, I find it interesting that the following biconditional holds: a space $X$ is $\Delta$-Hausdorff if and only if the loop-deletion procedure can be applied to every non-loop path to produce an injective path.
It's also worth pointing out that it is possible to prove the theorem directly without using the axiom of choice. Such a proof is due to R. Borger, but this paper is probably difficult to come by if you don't have access to a University library.
The improvement is modest but properties like the $\Delta$-Hausdorff (or weakly Hausdorff) property are often used in algebraic topology and are strictly weaker than the Hausdorff property. For instance, the weakly Hausdorff property is a convenient separation axiom in the category $CG$ of compactly generated spaces. The $\Delta$-Hausdorff property is to the category of $\Delta$-generated spaces as the weakly Hausdorff property is to compactly generated spaces.