A contractible space has trivial fundamental group

By [L1], there are a point $x_1\in X$ and a homotopy $H:X\times I\to X$ from ${\mathrm{id}}_X$ to the constant map $c_{x_1}$. Evaluating at the chosen basepoint gives a path $\lambda (t)=H(x_0,t)$ from $x_0$ to $x_1$.

Let $[\alpha ]\in {\pi }_1(X,x_0)$. Define $K:I\times I\to X$ by $K(s,t)=\{\begin{array}{ll}H(x_0,3st),& 0\le s\le \frac{1}{3},\\ H(\alpha (3s-1),t),& \frac{1}{3}\le s\le \frac{2}{3},\\ H(x_0,(3-3s)t),& \frac{2}{3}\le s\le 1.\end{array}$ The three pieces are continuous and agree on the seams $s=\frac{1}{3},\frac{2}{3}$ because $\alpha (0)=x_0=\alpha (1)$, so $K$ is continuous. Also $K(0,t)=x_0=K(1,t)$ for every $t$, so $K$ is a path homotopy relative to the endpoints between loops at $x_0$. At $t=0$ it is $c_{x_0}\ast \alpha \ast c_{x_0}$, and at $t=1$ it is $\lambda \ast c_{x_1}\ast \bar{\lambda }$. Hence $[c_{x_0}\ast \alpha \ast c_{x_0}]=[\lambda \ast c_{x_1}\ast \bar{\lambda }]\mathrm{.}$ Since $[c_{x_0}]$ is the identity by [L2], this gives $[\alpha ]=[\lambda \ast c_{x_1}\ast \bar{\lambda }]$.

Define $J:I\times I\to X$ by $J(s,u)=\{\begin{array}{ll}\lambda (3s(1-u)),& 0\le s\le \frac{1}{3},\\ \lambda (1-u),& \frac{1}{3}\le s\le \frac{2}{3},\\ \lambda ((3-3s)(1-u)),& \frac{2}{3}\le s\le 1.\end{array}$ The three pieces are continuous and agree on the seams, so $J$ is continuous. Also $J(0,u)=x_0=J(1,u)$ for every $u$, while $J(-,0)=\lambda \ast c_{x_1}\ast \bar{\lambda }$ and $J(-,1)=c_{x_0}$. Thus $[\lambda \ast c_{x_1}\ast \bar{\lambda }]=[c_{x_0}]$ by [L2], and step 2.1 yields $[\alpha ]=[c_{x_0}]$.

Every loop class at $x_0$ is therefore the identity, so ${\pi }_1(X,x_0)$ is trivial.