On Path-Connectedness Prove that if there is $x_0 \in X $, such that for any $x \in X$, there is a path
connecting $x$ to $x_0$, then $X$ is path connected.
I have trouble in provving it. Here is my attempt:
Fix a $x,y \in X$, then there exists $f_x,f_y:[0,1]\rightarrow X$ such that $f_x(0)=f_y(0)=x_0, f_x(1)=x, f_y(1)=y$.
Now define a new function $f:[0,1]\rightarrow X$ by $f(t)= f_x(1-2t)$ if $t\in [0,1/2)$ and $f(t)=f_y(2t-1)$ if $t\in [1/2,1]$. Then we have $f(0)=f_x(1-2(0))=f_x(1)=x, f(1)=f_y(2(1)-1)=f_y(1)=y.$ But I dont know how to prove that $f$ is contiinuous. I think we only need to prove the continuity at $t=1/2$. Actually I also want to ask is my chosen path the right one and if there is any alternative proof? Thank you so much!

As rghthndsd suggested, here it is:
The path you chose is simple and works. But any kind of concatenation of the paths $f_x$ and $f_y$ would work. To prove its continuity, you use the pasting lemma:
Let $X$ and $Y$ be topological spaces and suppose $X=A \cup B$ where $A$ and $B$ are open (or closed) subsets. Then a map $f:X \to Y$ is continuous if the restrictions $f|_A:A \to Y$ and $f|_B:B \to Y$ are continous.
Its not hard to prove it, but if you want I can elaborate.
