The proof is correct, although it would be preferable to explain why the zero ideal is a finite intersection of maximal right ideals a little more carefully. It is true that in an artinian poset of one sided ideals, a collection intersecting to zero must have a finite subset intersecting to zero, but it is not really done justice by saying “the set had a minimal element which is obviously (read “necessarily”) zero.”
My first thought was to take $R→⨁_M R/M$ [...] with $M$
running over all maximal left ideals, but I realized that I couldn't guarantee that this was a direct sum a priori.
Well, you can make that sum and it will be direct, nothing prevents you from doing so. But yes, the map you want to do (sending $r$ to its image in the quotient) doesn’t work. You would necessarily be working in $\prod_M R/M$ instead. Reducing it to a finite product makes it equivalent to a direct sum, and that was a very important reduction.
Is there a similar argument to prove that there are only finitely many maximal ideals in the noncommutative case?
Yes, the same argument (with noncommutative prime ideals) works. If a prime ideal contains a finite intersection of prime ideals, it necessarily contains their product, whence it contains at least one of the terms the product.
Suppose $R$ is a ring such that it can be written as $R=k_1\times \cdots \times k_n$ for some fields $k_i$. The ideals of any ring $A_1\times A_2$ are direct sums of the form $I_1\oplus I_2$, where $I_i$ is an ideal of $A_i$. If we think of $A_1\times A_2$ as a module over itself, then its submodules are precisely the ideals, in particular $A_1\times A_2 = A_1\oplus A_2$ as a module over itself. Applying this to $R$, we have the module $R=k_1\oplus\cdots\oplus k_n$. Since a field is a simple module over itself, the module $R$ is a finite direct sum of simple modules, and therefore a semisimple module over itself. This means it is semisimple.
Conversely, suppose $R$ is semisimple and of the form $R=L_1\oplus\cdots\oplus L_n$. Then we can write the unit element $1$ as a finite sum $1=e_1 + \cdots + e_n$, where $e_i\in L_i$. The important thing to notice here is that $e_ie_j = 0$ if $i\neq j$ ($L_iL_j \subseteq L_i\cap L_j = 0$), and $e_i^2=e_i$ ($e_i = 1e_i = (e_1 + \cdots + e_n)e_i = e_1e_i + \cdots + e_ne_i$, where all but $e_i^2$ are zero by the previous observation). We can therefore write $L_i = \langle e_i\rangle = \{ re_i\ |\ r\in R\}$ (since, again, we have the direct sum decomposition, $re_i=0$ whenever $r\not\in L_i$. Furthermore, since we know that each $L_i$ is an ideal, it is additively closed and contains $0$, but we also have $(re_i)(se_i) = (rs)e_i^2 = (rs)e_i$ for any $r, s\in R$, so it is also multiplicatively closed, and $e_i$ is its unit element ($e_i(re_i) = re_i^2 = re_i$ for any $re_i \in L_i$). Therefore, $L_i$ is a subring of $R$.
We may now start quotienting out the $L_i$. First, some notation. Let $R_j = R/(L_1 \oplus \cdots \oplus L_{j-1})$, the quotient with the subrings from $1$ to $j-1$ quotiented out, and $R^m_j = R_j/(L_m\oplus\cdots\oplus L_n)$, the subrings starting from $m$ quotiented out. First, $L_n = R_n$, and it is a simple module both over $R$ and itself, so it is a simple commutative ring (a field). Next, $L_{n-1} = R_{n-1}/R_j^n$, which is (by the same arguments) a field. This holds for each $L_i = R_i/R_i^{i+1}$, so we conclude that $R$ is indeed a finite direct product of fields.
Best Answer
You mean semisimple as a module over itself. Examples include
It follows that any ring which is a finite product of matrix rings over division rings is semisimple. By the Artin-Wedderburn theorem, this exhausts all possibilities. In particular, $\mathbb{Z}$ is not semisimple: it is not semisimple as a module over itself because it has nontrivial submodules, such as $2 \mathbb{Z}$, which don't have complements.
In practice, a common source of examples of semisimple rings come from finite groups: if $k$ is a field and $G$ is a finite group of order not divisible by the characteristic of $k$, then the group ring $k[G]$ is semisimple by Maschke's theorem: equivalently, the category of $G$-representations over $k$ is semisimple.
See this blog post for a longer discussion with proofs.