(1.32) Observe that $Deck(\tilde{X},u)=\pi_1(X,x)$. This group acts on $\tilde{X}$. Our strategy is as follows:

1. We will first prove that this action is properly discontinuous.
2. Then we will form the quotient $Y=\tilde{X}/H$. Since $\tilde{X}$ is simply-connected and since the action is properly discontinuous, we deduce that $\pi_1(Y,[\tilde{x}]_H)=H$ (where $\tilde{x}\in\tilde{X}$ is a basepoint and $[\tilde{x}]_H\in Y$ denotes its orbit under $H$).
3. Finally, we need to check that the projection map $p\colon Y\to X$ is a covering map; the projection $p$ is the map defined as follows: a point $y\in\tilde{X}$ defines an equivalence class $[y]_H\in\tilde{X}/H$ and an equivalence class $[y]_X\in\tilde{X}/Deck(X,x)=X$; we define $p([y]_H)=[y]_X$.

(4.44) Once we have proved these facts, we obtain a covering space $p\colon Y\to X$ with fundamental group isomorphic to $H$. It is an exercise to think about why $p_*\pi_1(Y,[\tilde{x}]_H)$ is precisely the subgroup $H\subset\pi_1(X,x)$, rather than just some subgroup of $\pi_1(X,x)$ isomorphic to $H$.

(6.04) First, we will prove that the deck group of any covering space $p\colon Y\to X$ acts properly discontinuously. Pick a point $y\in Y$ and set $x=p(y)$. To prove that the action of $Deck(Y,p)$ is properly discontinuous, we need to find a neighbourhood $V$ of $y$ such that $F(V)\cap V$ is empty for any $F\in Deck(Y,p$ not equal to the identity. To that end, pick an elementary neighbourhood $U$ containing $x$ and let $V$ be the elementary sheet in $Y$ over $U$ containing $y$.

(8.12) Take $g\in Deck(Y,p)$. We want $Vg\cap V=\emptyset$ (writing our action on the right). Because $g$ is a covering transformation, $Vg$ is another elementary sheet for $p$ living over $U$. We have two possibilities: either $Vg=V$ or else $Vg\cap V=\emptyset$. If $Vg=V$ then $yg=y$ (as $y$ is the unique preimage for $y$ under $p$ in $V$), so $g$ fixes a point and hence agrees with the identity (by the uniqueness theorem for covering spaces). Therefore $Vg\cap V=\emptyset$ unless $g=id_Y$.

(11.02) Finally, we show that if a group $G$ acts properly discontinuously on a space $Z$ and $H\subset G$ is a subgroup then the quotient map $p\colon Z/H\to Z/G$ is a covering map. (To relate this to point (3) mentioned earlier, take $Z=\tilde{X}$, $G=\pi_1(X,x)$ so $X=Z/G$.)

(11.50) We will write $[z]_H\in Z/H$ and $[z]_G\in Z/G$ for the respective equivalence classes of a point $z\in Z$. The map $p\colon Z/H\to Z/G$ is $p([z]_H)=[z]_G$. We need to check that $p$ is well-defined and continuous. If $[z]_H=[z']_H$ then there exists $h\in H$ such that $z=z'h$. Since $H\subset G$ this implies that $[z]_G=[z']_G$. There is a quotient map $q_G\colon Z\to Z/G$ which is continuous by definition of the quotient topology on $Z/G$. There is also a quotient map $q_H\colon Z\to Z/H$, which is again continuous. By the video on continuous maps out of quotient spaces, $p\colon Z/H\to Z/G$ is the unique map we need to get $q_G=p\circ q_H$ and is therefore continuous.

(14.53) To prove that $p$ is a covering map, we use the fact that $q_G\colon Z\to Z/G$ is a covering map. We have elementary neighbourhoods $U\subset Z/G$ and local inverses for $q_G$ defined over $U$. If we compose these local inverses with the map $q_H$, we get local inverses to $p$ defined on the same elementary neighbourhoods. This shows that $p$ is a covering map.