A proof of: $S^n$ is simply connected when $n \geq 2$ There is an exercise of "Kosniowski" about a way to get a proof of the simple connection of $S^n$, when $n \geq 2$. I cannot understand a point.
Let $n \geq 2$. Due to the fact that $S^n$ is connected, locally pathwise connected and semi-locally pathwise connected, there is a universal covering $p \colon E \to S^n$. Then $E$ is simply connected.
Let $f$ be a continuous map $D^n \to S^n$ such that $f(\partial D^n)$ is a singleton (for instance, $f$ could map every point of $D^n$ into a fixed point of $S^n$). By a lifting criterion, being $E$ simply connected, there is a continuous map $f' \colon D^n \to E$ such that $p \circ f' = f$.
Now, suppose that there is a continuous section of $p$, i.e. a continuous map $f'' \colon S^n \to E$ such that $p \circ f'' = 1_{S^n}$. Then, being $\pi_1$ a functor, $\pi_1(p) \circ \pi_1(f'')=1_{\pi_1(S^n)}$. Notice that $\pi_1(f'')$ maps everything into the neutral element of $\pi_1(E)$, because $\pi_1(E)$ is the trivial group. Then $1_{\pi_1(S^n)}$ maps everything into the neutral element of $\pi_1(S^n)$, and this implies that $\pi_1(S^n)$ is trivial, being $1_{S^n}$ injective. Then we are done, if such a section exists.
How can I prove the existence of such an $f''$? In the exercise I read that you need to use $f'$ to define $f''$, but I do not know how.
 A: 
Let $f$ be a continuous map $D^n \to S^n$ such that $f(\partial D^n)$ is a singleton (for instance, $f$ could map every point of $D^n$ into a fixed point of $S^n$). For a lifting criterion, being $E$ simply connected, there is a continuous map $f' \colon D^n \to E$ such that $p \circ f' = f$.

You don't need to use the fact that $E$ is simply connected to prove the existence of such a map $f'$. This map exists because of the homotopy lifting property of coverings, given that you can consider the map $f$ as a homotopy between $f|_{\partial D^n}$ and $f|_{\{0\}}$, seen as maps $S^{n-1}\to S^n$.
Now the point is that you can't take just any $f'$ for which $p \circ f' = f$. You also need that $f'(\partial D^n)$ is a single point in order to define $f''$ later. 
The reason why there is such an $f'$ is because the homotopy lifting property lets you impose the value of the lift at one extremity. Therefore, since $f(\partial D^n)$ is a single point, you can lift the homotopy there by choosing an arbitrary point in $E$ and setting $f'(\partial D^n)=$ that point, then lift the rest of the homotopy.

How can I prove the existence of such an $f''$?

I assume known that $S^n\setminus \{\text{one point}\}$ is homeomorphic to $\mathbb{R}^n$ and hence to an open $n$-dimensional ball.
One way to say this is to consider the topological quotient map $D^n\to D^n/\sim$ where interior points are alone in their equivalence class and all boundary points are identified with each other: $D^n/\sim$ is actually homeomorphic to $S^n$. 
Now take $f$ to be the map $D^n\to S^n$ induced by the above projection. The requirement that $\partial D^n$ should be mapped to a single point is met. 
Since $f$ is a homeomorphism on the interior it has an inverse $g$.
Define $f''=f'\circ g$ on the sphere $S^n\setminus \{\text{one point}\}$, and set $f''(\{\text{that one point}\})$ to be equal to $f'(\partial D^n)$. This is where you crucially need $f'(\partial D^n)$ to be one point, so that $f''$ is well-defined and continuous.
