As written, this answer is a sort of intuitive yes. In fact, let's make the path almost everywhere smooth. All the necessary intuition is to believe that if we have three points with prescribed tangent vectors at each, then there is a smooth path between them (sort of - I suppose there's a bit more). There are a few ways to see this: you might restrict yourself to the plane containing those three points and perform a sort of quintic Hermite approximation (or closer, a Bezier curve-like approximation). It might be a bit better if you actually straighten out the path in a small neighborhood around each of the points (small enough to not intersect the other neighborhoods), which you can justify with a bump function style argument. This isn't necessary, but perhaps makes it easier to visualize, as then the assertion is that you can 'smoothly twist' a line to go through any point? (I'm a bit uncertain of your familiarity with these things).
Then the idea would be to start with a straight line segment through the origin, which is clearly 'differentiable' at the origin. Then smoothly connect the line segment to $x_1$. And then to $x_2$. And so on.
If you're a generalized path sympathizer, then you might let $\gamma$ traverse the path from $x_n$ to $x_{n+1}$ at time $t \in [n, n+1]$, with the idea that traversing paths back to back is still a path. Or perhaps you're a stickler, and then you'll have to do this in something like at time $t \in [1 - 2^{-n}, 1 - 2^{-(n+1)}]$, and traversing the $n$th path at $2^n$th rate (still a path, though). If in addition, we bound the $n$th path in some decreasing neighborhood of the origin (which should exist as $x_n \to 0$), I believe we should even be able to say that $\gamma(1) = 0$.
The idea here is that you can sort of piecemeal your way through so that the path is smooth on $t \in (0,1)$, with $\gamma (0) = 0$. And with this particular construction, $\lim_{t \to 0} \gamma'(t)$ is defined, and gives the direction of the original line segment. And the fact that $x_n \to 0$ should mean that our path should descend to $0$ as well, although there's not a chance in the world that it will be smooth at $t = 1$.
The resulting path will be a little bit wonky, perhaps, but this seems reasonable in principle.