Two definitions of simply connected I'm showing that the first definition here implies the second (the other implication is obvious). My thoughts: Let $p,q$ be two paths in the space $X$. Then since $X$ is path connected there are two paths $f,g$ connecting the two endpoints of $p,q$ respectively. The concatenation of $q.g.p.f$ is a map $S^1 \to X$ and by assumption it extends continuously to some $F : D^2 \to X$. 
Here is where I'm stuck. May be it's not even the right ideas I have. Can you please help me?  
 A: As in the second definition, take two paths $p:[0,1]\rightarrow X$ and $q:[0,1]\rightarrow X$ that have the same start and end point, let $\tilde{q}$ denote the reverse path, ie the one that goes the other way, so $p(1)=\tilde{q}(0)$. The concatenated path $p\cdot \tilde{q} :[0,1]\rightarrow X$ satisfies $p\cdot \tilde{q}(0)=p\cdot \tilde{q}(1)$, so is equivalent to a map $p\cdot \tilde{q}: S^1\rightarrow X$. Intuitively, this map does $p$ on one half of the circle and $\tilde{q}$ on the other half.
The first definition now tells us that $p\cdot \tilde{q}$ extends to a map $F:D^2\rightarrow X$, where we can think of $p$ mapping from one side of the boundary and $q$ mapping from the other. Now define a homotopy from $p$ to $q$ by "moving across" the disk $D^2$ using $F$. To write this down succinctly we can write $F$ as
$F:[0,1]\times[0,1]\rightarrow X$
with $F(0,t)=F(1,t)$ for all $t\in[0,1]$. Then define the homotopy $F_t$ from $p$ to $q$ by $F_t(s)=F(s,t)$. This is definitely continuous, it fixes $\{0,1\}$ (so is a homotopy rel $\{0,1\}$) and it satisfies $F_0=p$ and $F_1=q$ by construction.
A: Assume WLOG that the paths have  length $1$. Consider $w:[0,1]\rightarrow \mathbb{S}^1$ send $x$ to $(\cos(\frac{x}{2\pi}),\sin(\frac{x}{2\pi}$)). Verify that $w$ is an identification map. Consider $f:\mathbb{S}^1\rightarrow X$ that sends $(m,n)$ to $(-q+p)[arctan(\frac{m}{n})]$. Verify that $fw=-q+p$. Since $w$ is an identification map and $-q+p$ is continuous. Therefore $f$ is a continuous function. Thus, there exists a continuous function $F:D^2\rightarrow X$ such that $F|\mathbb{S}^1=f$. Now remember that $D^2\cong [0,1]\times[0,1]$. Let $h:D^2\rightarrow [0,1]\times [0,1]$ be a homeomorphism. Verfiy that $h^{-1}F$ is a homotopy from  $p$ to $q$.
