Why are these all the indecomposable projective modules?

Let $A$ be a finite dimensional associative algebra with unit over a commutative field $k$. Suppose that $M$ is a finitely generated left module. Denote by $I(M)$ the injective hull of $M$ and by $P(M)$ the projective cover of $M$. Suppose that $\{S_i\}_{1\leq i\leq n}$ is the set of all nonisomorphic simple $A$-modules (why is it finite?). Then $\{P(S_i)\}_i$ are all nonisomorphic indecomposable projective modules and $\{I(S_i)\}_i$ all nonisomorphic indecomposable injective modules, why is this true?

I'm also wondering why there exist the projective covers?

If I call $\mathrm{top}\;M$ the factor module $M/\mathrm{rad}\;M$ where $\mathrm{rad}\;M$ is the radical of $M$, then why $\mathrm{top}\;P(S_i)\cong\;S_i\cong\mathrm{soc}\;I(S_i)$?

And why $I(S_i)\cong\mathrm{Hom}(\mathrm{Hom}_A(P(S_i),A),k)$ ?

I know that these are a lot of questions, but it makes more sense putting them together in one unique question.

Let me start with the indecomposable projective modules. I claim that they are all summands of $A$ (Theorem 20.2): Indeed let $P$ be an arbitrary finitely generated indecomposable projective module. Then $A^n \twoheadrightarrow P$. This epimorphism splits, so Krull-Remak-Schmidt gives that $P$ is a direct summand of $A$. So there are as many (f.g. indec.) projective modules as isomorphism classes of such summands of $A$. Since $A$ is finite-dimensional these are finitely many.
The next step is to set up a bijection from indecomposable projectives to simples (always isoclasses). This is given by sending $P$ to $P/\operatorname{rad}(P)$ (Theorem 20.3). This is surjective because given some simple $S$, there is a surjection $A\to S$. Hence $P_i\to S$ for some $i$, surjective because $S$ is simple. Now since $P_i$ is local (to check), the kernel is $\operatorname{rad}(P_i)$. Injectivity follows e.g. from Schanuel's lemma or from unicity of projective covers as it is done in the lecture notes.
Existence of projective covers (Theorem 20.5): Let $M$ be some f.g. module. Then there exists some epimorphism $p:P\twoheadrightarrow M$ (e.g. $P=A^n$). Now choose $P$ to be minimal with respect to the number of direct summands. We want to show it is a projective cover. One characterisation of projective covers (for f.g. modules) is to prove that $\ker(p)\subseteq \operatorname{rad}(P)$. Then there is a direct summand of $P$ in $\ker(p)$ contradicting the minimality of $P$.
That $I(S_i)\cong \operatorname{Hom}(\operatorname{Hom}_A(P(S_i),A),k)$ follows from the facts that $\operatorname{Hom}_A(-,A)$ is a duality mapping projective $A$-modules to projective $A^{op}$-modules and $\operatorname{Hom}(-,k)$ is a duality mapping projective $A^{op}$-modules to injective $A$-modules. Now carefully sending the epimorphism $P(S_i)\to S_i$ through the dualities and using uniqueness of injective envelopes you get the result.