Let $(X,*)$ be a pointed topological space.
Let $F(X,k)=\{(x_1,\cdots,x_k)\in X^k\mid \forall i\neq j: x_i\neq x_j, \}$.
Let $F(X,k)/S_k$ be the $k$-th unordered configuration space.
Is there an inclusion $F(X,k)/S_k\to F(X,k+1)/S_{k+1}$ for each $k\geq 1$?
Note that $[x_1,\cdots,x_k]\mapsto [x_1,\cdots,x_k,*]$ is not well-defined.
This is true when $X=\mathbb{R}^n$. I want to try $X=S^n, \mathbb{R}P^n,\mathbb{C}P^n$...
Is there other way?
With your definitions, assuming you mean the configuration space of distinct points, and that the inclusions need to be compatible with the actual locations of the points in some way, there is not such a map for homotopy reasons.
I will write $C(X,k)$ for the unordered configuration space.
Let $X$ be the circle $S^1$. Then $F(X,k)$ is homotopy equivalent to a union of $(k-1)!$ copies of the circle, and $C(X,k)$ is a single circle. The quotient map to $F(X,k)$ is a covering map of degree $k!$; each component of $F(X,k)$ is maps to $C(X,k)$ with degree $k$.
There is an inclusion $C(X,1)$ to $C(X,k)$ for each $k$; it's easy to construct it explicitly, and it is a map of degree $k$ as a map from one circle to another.
There is not, however, a map from $C(X,2)$ to $C(X,3)$ that is compatible with the point positions. Such a map would have to be a covering map of "degree $3/2$", which is impossible.
I got a request for some more details on the reasoning. $F(S^1,k)$ breaks into connected components according to the cyclic ordering of the $k$ points $x_1,\dots,x_k$. There are $(k-1)!$ different cyclic orderings of $k$ points, giving the $(k-1)!$ connected components. On each component, we can take the position of $x_1$ as one of the coordinates, and then keep track of the successive differences between adjacent points (in the cyclic order) as the remaining coordinates. The successive differences are a choice of $k-1$ positive reals whose sum is less than one, which is a $(k-1)$-dimensional simplex. Thus this component of $F(S^1,k)$ is homeomorphic to a circle cross a simplex, which is homotopy equivalent to a circle, as claimed.
To find $C(S^1,k)$, we take the quotient of $F(S^1,k)$ by the symmetric group $S_k$. Some of the permutations will identify the different components of $F(S^1,k)$, but there is also a $k$-cycle in the given cyclic order, which takes that component to itself. As a result, $C(S^1,k)$ is connected, and is homeomorphic to a non-trivial simplex bundle over $S^1$. (For instance, $C(S^1,2)$ is a Möbius strip.) $C(S^1,k)$ is still homotopy-equivalent to a circle. The generator of $\pi_1$ of a component of $F(S^1,k)$ is a loop that takes $x_1$ all the way around the circle to itself, while the generator of $\pi_1$ of a component of $C(S^1,k)$ cyclically permutes the $x_i$.
