Why is the Pontraygin dual a locally compact group?
I'm not sure whether directly trying to show that every open cover of the given set has a finite subcover leads to anything. I remember running into one wall after the other for a long time trying to prove that $\hat{G}$ is locally compact too.
I think it's best to take a small detour. Consider the space $X = (S^1)^G$ of all maps $G \to S^1$. We endow it with the product topology, or equivalently the topology of pointwise convergence. By Tíkhonov's theorem, $X$ is compact. In $X$, consider the subspace $\tilde{G}$ of characters (continuous or not) on $G$. Since
$$\tilde{G} = \bigcap_{x,y \in G} \{ f\in X : f(x)f(y) = f(xy)\},$$
we see that $\tilde{G}$ is a closed subspace of $X$, hence compact. Also, for every $S\subset G$ and $\epsilon > 0$, the set
$$A(S,\epsilon) = \bigl\{ f\in X : \bigl(\forall x\in S\bigr)\bigl(\lvert f(x) - 1\rvert \leqslant \epsilon\bigr)\bigr\}$$
is closed in $X$, hence compact. In particular, for a neighbourhood $U$ of the identity in $G$, and $\epsilon > 0$, the set
$$V(U,\epsilon) = \tilde{G} \cap A(U,\epsilon)$$
is a closed subset of $X$, hence compact.
Now we note that $V(U,\epsilon) \subset \hat{G}$ for small enough $\epsilon$:
For simplicity, we use the arc-length distance on $S^1$, not the Euclidean distance. Assume $\epsilon < \frac{\pi}{2}$. Starting with $U_0 = U$, choose a sequence $(U_n)$ of neighbourhoods of the identity in $G$ such that $U_{n+1}\cdot U_{n+1} \subset U_n$ for all $n \in \mathbb{N}$, and set $\epsilon_n = 2^{-n}\cdot \epsilon$. For $x \in U_{n+1}$ and $\chi \in V(U_{n}, \epsilon_{n})$, we have $\lvert \chi(x) - 1\rvert \leqslant \epsilon_n$ since $x\in U_n$, and $\lvert \chi(x)^2 - 1\rvert = \lvert \chi(x^2) - 1\rvert \leqslant \epsilon_n$, since $x^2 \in U_n$. The latter implies $\lvert \chi(x) - 1\rvert \leqslant \epsilon_n/2 = \epsilon_{n+1}$ since $\epsilon_n < \frac{\pi}{2}$. Hence $V(U_n,\epsilon_n) \subset V(U_{n+1}, \epsilon_{n+1})$.
In particular $V(U,\epsilon) \subset V(U_n,\epsilon_n)$ for all $n$, and since for every neighbourhood $W$ of $1$ in $S^1$ there is an $n\in\mathbb{N}$ such that $\{ z : \lvert z-1\rvert \leqslant \epsilon_n\} \subset W$, we see that every $\chi \in V(U,\epsilon)$ is continuous at the identity of $G$, and as a homomorphism between topological groups, it is continuous everywhere.
We saw above that $V(U,\epsilon)$ is compact in the topology of pointwise convergence, but what we need is that $V(U,\epsilon)$ is compact in the topology of compact convergence (or locally uniform convergence, since $G$ is locally compact).
However, if we look at the proof of continuity above a little closer, we see that we proved more than just $V(U,\epsilon) \subset \hat{G}$. In fact, we proved that $V(U,\epsilon)$ is a uniformly equicontinuous family. For all $\chi \in V(U,\epsilon)$ and $x,y\in G$ with $x^{-1}y \in U_n$, we have
$$\lvert \chi(y) - \chi(x)\rvert = \lvert \chi(x)^{-1}\chi(y) - 1\rvert = \lvert \chi(x^{-1}y) - 1\rvert \leqslant \epsilon_n.$$
Now we can invoke a theorem of Ascoli-Bourbaki,
Let $X$ be a topological space and $Y$ a uniform space. Let $H \subset C(X,Y)$ be equicontinuous. Then on $H$, the uniform structures of compact convergence and of pointwise convergence coincide.
In particular, the topologies of compact convergence and of pointwise convergence coincide on $H$.
And thus we see that $V(U,\epsilon)$ is compact in the topology of compact convergence.
Finally, if $U$ is a compact neighbourhood of the identity in $G$, then $V(U,\epsilon)$ is a neighbourhood of the identity in $\hat{G}$, and we see that $\hat{G}$ with the topology of compact convergence is locally compact.