A contractible space is path connected. Let the space be $X$ and $\rm{id} \simeq x_0$  where $\rm{id}$ is the identity map on $X$ and $\rm x_0$ is some fixed point in $X$.   How do I show that for any two points $a,b \in X$ there is a continuous path $f:[0,1] \to X$ such that $f(0) = a, f(1) = b$?
 A: If $X$ is contractible, say to $x_0\in X$, there is a continuous function
$$
f: I\times X\to X,
$$
where $I=[0,1]$, such that $f(0,x)=x$ and $f(1,x)=x_0$, for all $x\in X$.
Now let $a,b\in X$, and define the path $\gamma: I\to X$ as
$$
\gamma(t)=\left\{\begin{array}{lll}
f(2t,a) & \text{if} & t\in[0,1/2], \\
f(2-2t,b) & \text{if} & t\in[1/2,1].
\end{array}\right.
$$
Clearly $\gamma$ is continuous (Pasting Lemma), and also $\gamma(0)=a$ and $\gamma(1)=b$.
A: Consider a homotopy between the identity map on $X$ and the point $x_0$.  Choose a point $a$ and using this homotopy find a path between $a$ and $x_0$.  
A: Take any two points $p$ and $q$ and a homotopy $F(x,t)$ such that $F(x,0)=x$ and $F(x,1)=z$ for some fixed point $z$ . Then $F(p,t)$ as a function in $t$ defines a path from $p$ to $z$ , and $F(q,t)$ defines a path from $q$ to $z$ , so that all three points $p$ , $q$ and $z$ are in the same path connected component.
A: I will add my answer here to give the readers of this question a bit more explanation as to how one could prove this.
Proof: Suppose that $X$ is contractible. Choose any two points $a, b \in X$. We will construct a path between $a$ and $b$. 
Since $X$ is contractible, the identity map $\operatorname{id}_{X} : X \to X$ defined by $\operatorname{id}_X(x) = x$ for all $x \in X$, is null-homotopic. That means that $\operatorname{id}_X$ is homotopic to a constant map $c_p : X \to X$ defined by $c_p(x) = p$ for all $x \in X$.  Let $H : X \times I \to X$ be the homotopy between $\operatorname{id}_X$ and $c_p$. Then $H$ is continuous and $$H(x, 0) = \operatorname{id}_X(x)= x \ \ \ \ \ \forall x \in X$$ $$H(x, 1) = c_p(x) = p \ \ \ \ \ \forall x \in X$$
Now the idea to construct the path from $a$ to $b$ is the following, first we define a path $\varphi : I \to X$ from $a$ to $p$ by $\varphi(t) = H(a, t)$. Note that $\varphi(0) = H(a, 0) = a$ and $\varphi(1) = H(a, 1) = p$  so $\varphi$ is indeed a path from $a$ to $p$. 
Then we define a path $\psi : I \to X$ from $b$ to $p$  by $\psi(t) = H(b, t)$. Again note that $\psi(0) = H(b, 0) = b$ and $\psi(1) = H(b, 1) = p$  so $\psi$ is indeed a path from $b$ to $p$. 
Now we wish to combine these two paths and traverse them 'at twice speed' to get a path $\gamma : I \to X$ that goes from $a$ (through $p$) to $b$. But right now we have a path from $b$ to $p$, and we need a path from $p$ to $b$ to make this idea work. 
To remedy this problem we define what's called a 'reverse path' of the path $\psi$ which goes from $b$ to $p$. Define $\psi^{-1} : I \to X$ by $\psi^{-1}(t) = \psi(1-t) = H(b, 1-t)$. Then you can check that $\psi^{-1}$ is a path from $p$ to $b$. 
Now finally define $\gamma : I \to X$ by $$\gamma(t) = 
\begin{cases}
\varphi(2t) \ \ \ \ \  \ \ \ \ \ \ \ \ \ \  \ \  \ 0 \leq t \leq \frac{1}{2}\\
\psi^{-1}(2t-1) \ \ \ \ \ \  \ \frac{1}{2} \leq t \leq 1
\end{cases}
$$
You can check that $\gamma(0) = a$ and $\gamma(1) = b$ and that $\gamma$ is continuous by the gluing lemma since $\varphi(1) = \psi^{-1}(0)$.Thus $\gamma$ is a path from $a$ to $b$. Hence $X$ is path-connected as desired. $\square$
