If your main concern is to have a reference to quote, then 1) is Theorem III.1.1 and 2) is essentially Corollary III.1.2 or Lemma III.2.1 in Davidson's book, as mentioned by Mike Jury.

This answer just aims at giving a direct (not using the larger compact case as in Davidson's book) self-contained and elementary operator algebraic approach. Hopefully this is correct...

Your first statement can be seen as the finite-dimensional case of the structure of a type I von Neumann algebra which somehow generalizes Artin-Wedderburn over $\mathbb{C}$. Your second statement is the root of Brattelli diagrams. Both can be proved by elementary manipulations of the projections of the algebra.

Note that if a finite-dimensional $*$-subalgebra $A$ of $B(H)$ does not contain $\mbox{Id}_H$, still the unital $*$-algebra $A\oplus \mathbb{C}\mbox{Id}_H$ is spanned by its projections by functional calculus (or just diagonalization of normal operators), so $A$ is also spanned by its projections (projections of $A\oplus \mathbb{C}\mbox{Id}_H$ are $p\oplus 0$ and $-p\oplus \mbox{Id}_H$ for projections $p$ of $A$). Whence the joint of all the projections of $A$ is a unit for $A$. So a finite-dimensional $*$-subalgebra of $B(H)$ is always a unital von Neumann algebra. Up to reducing $H$, we can assume that this unit is $\mbox{Id}_H$.

1) If $A$ is a finite-dimensional $*$-subalgebra of $B(H)$, then $A$ is unitarily equivalent to $\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$.

In short: a finite-dimensional von Neumann algebra must be of type $\rm{I}$, whence unitarily equivalent to $\sum_{\alpha}^\oplus A_\alpha \otimes B(H_\alpha)$ with $A_\alpha$ Abelian (Theorem I.27 in Takesaki I). In the finite-dimensional case, this gives the assertion.

Here is how the above goes in the finite-dimensional case directly. First, considering the minimal projections of the center $A\cap A'\simeq \mathbb{C}^r$ which decompose $A$ into blocks, we can restrict to the factor case $A\cap A'=\mathbb{C}1$. By compactness of the unit sphere of $A$, there are finitely many Murray-von Neumann equivalence classes of projections since two close enough projections ($\|p-q\|<1$) must be homotopic, whence equivalent. Therefore, there are minimal projections in particular. Since the order on projections is total in a factor, all the minimal projections are equivalent. Finally, take $p_1\oplus \ldots \oplus p_n$ a maximal orthogonal set of minimal projections. By maximality, we must have $p_1\oplus \ldots \oplus p_n=1$. Since these minimal projections must be pairwise equivalent, the partial isometries implementing these equivalences yield a unitary equivalence between $A$ and $M_n(\mathbb{C})\otimes p_1 Ap_1$. And by minimality of $p_1$, $p_1Ap_1\simeq \mathbb{C}$.

Decomposing the range into blocks, your second question follows from the following fact.

2) If $A=\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$, and if $\pi:A\longrightarrow B(H)$ is a $*$-homomorphism, then $\pi$ is unitarily equivalent to an orthogonal direct sum of possibly zero diagonal embeddings $x_j\longmapsto x_j\otimes 1_j$ where $x=(x_1,\ldots,x_r)\in A$.

Without loss of generality, we can assume that $\pi$ is unital, up to reducing $H$.

The key point is that a unital $*$-homomorphism $\pi:A\longrightarrow B(H)$ sends projections to projections (self-adjoint idempotents), and partial isometries to partial isometries.

First, if $A=\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$, then the corresponding orthogonal decomposition of the unit $1=\bigoplus p_j$ is sent to an orthogonal decomposition of the unit of $B(H)$. This allows us to restrict to the case $A=M_n(\mathbb{C})$, and to assume $\pi$ unital again.

Now the matrix units $e_{ij}$ with $1$ in $(ij)$-th position and $0$ elsewhere can be interpreted as partial isometries implementing the Murray-von Neumann equivalence of the projections $e_{ii}$ and $e_{jj}$. This is preserved under the unital $*$-homomorphism $\pi$. It follows that $\pi$ is, up to unitary equivalence, a diagonal embedding.

The case $A=M_2(\mathbb{C})$, with weaker assumptions, is used and explained with somewhat lengthy details in this MSE answer. It works the exact same way for $A=M_n(\mathbb{C})$.

Algebras of Linear Transformations(uregina.ca/~farenick/alt_contents.html ) although I don't have a copy at hand to check. The issue is to check, tacitly or explicitly, that such an algebra is semisimple in the ring-theoretic sense, then as you note Artin-Wedderburn gives you the rest $\endgroup$ – Yemon Choi Jul 24 '13 at 21:44