Path component of the direct limit of topological spaces. We consider a direct limit of a tower $X_1\subset\cdots\subset X_n$ of spaces, where each $X_n$ is a subspace of $X_{n+1}$. The direct limit is $X_{\infty}:=\cup_n X_n$ endowed with the topology $\mathcal{T}_{\infty}$ defined as follows: $U\subset X_{\infty}$ is open if and only if $\forall n\in \Bbb{N},\quad U\cap X_n$ is open on $\mathcal{T_n}.$
I don't have any idea for the following question (I can write the definition and so on but I am stuck ):

Is it true if $x$ and $y$ are in the same path component of $X_{\infty}$ then there are also on same path component of $X_{n}$ for $n$ big enough?

Defintion path component of a topological space $Y$ are the equivalence classes of the equivalence relation on $Y$ defined by $x\sim y$ if and only if there exist a path connecting $x$ to $y$
edit It's assumed that $X_{\infty}$ is Hausdorff and each $X_i$ is closed in $X_{i+1}$ for all $i\ge 1$.
 A: This is true assuming each $X_n$ is $T_1$ (in particular, assuming $X_\infty$ is Hausdorff, as you are assuming); I don't know whether it is true without this assumption.  It suffices to show that if $f:[0,1]\to X_\infty$ is continuous, then the image of $f$ is contained in $X_n$ for some $n$.  The image of $f$ is compact, so it suffices to show that any compact subset $K\subseteq X_\infty$ is contained in $X_n$ for some $n$.
To prove this, suppose $K$ is not contained in $X_n$ for any $n$.  Then for each $n$, we can choose a point $a_n\in K\setminus X_n$.  Let $A$ be the set of all these points $a_n$.  Note that the intersection of $A$ with each $X_n$ is finite, and hence closed in $X_n$ (since $X_n$ is $T_1$).  It follows that $A$ is closed, and so it is compact since $K$ is compact and $A\subseteq K$.  But the same argument also shows that any subset $B\subseteq A$ is closed as well, so $A$ is discrete.  Since $A$ is discrete and compact, it must be finite.  But any finite subset of $A$ is contained in $X_n$ for some $n$ and $A$ is not contained in $X_n$ for any $n$ since $a_n\in A\setminus X_n$.  This is a contradiction.
