A key part of the story here is that a $C^*$-algebra whose self-adjoint elements all have finite spectrum must be finite-dimensional. Martin Argerami gave a reference which implies this result by way of several more general results in the context of Banach algebras. However, I decided to write up an account specifically for the $C^*$-algebra case in order to take advantage of the simplifications this context has to offer.
Claim 1: Suppose $A$ is a unital $C^*$-algebra all of whose self-adjoint elements have one point spectrum. Then $A$ is one-dimensional.
Sketch: Use functional calculus to show every self-adjoint element is a multiple of $1_A$, then use the fact that every $C^*$-algebra is the span of its self-adjoint part.
Claim 2: Let $A$ be a $C^*$-algebra in which every self-adjoint element has a finite spectrum and let $p$ be a nonzero projection in $A$ such that the "corner algebra" $pAp$ is not one dimensional. Then, one can write $p=p_1+p_2$ where $p_1$ and $p_2$ are nonzero, orthogonal projections.
Sketch: Working inside $pAp$, whose unit is $p$, the previous claim gives a self-adjoint element whose spectrum is a finite set with at least two elements. Now use functional calculus.
Claim 3: Let $A$ be unital $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then there exist nonzero, pairwise orthogonal projections $p_1,\ldots,p_n$ summing to $1_A$ such that each of the corner algebras $p_iAp_i$ is one-dimensional.
Sketch: If $A$ is not one-dimensional, subdivide $1_A$ into two projections. If either of the new projections does not determine a one-dimensional corner, subdivide again. This process must eventually terminate and give a decomposition of the desired type. Otherwise, we would have an infinite collection of pairwise orthogonal projections in $A$ which could then be used to embed a copy of $c_0(\mathbb{N})$ into $A$ leading to self-adjoint elements of infinite spectrum, contrary to assumption.
Claim 4: Let $p$ and $q$ be nonzero, orthogonal projections in a $C^*$-algebra $A$ satisfying that $pAp$ and $qAq$ are one-dimensional. Then $qAp$ and $pAq$ are either both one-dimensional or both zero.
Proof: Since $(qAp)^* = pAq$, it suffices to look at $qAp$. Fix any nonzero $a \in qAp$. By the $C^*$-identity, $a^*a \in pAp$ and $aa^* \in qAq$ are nonzero. Since $pAp$ is one-dimensional, up to rescaling $a$, we may assume $a^*a=p$. But then $a$ is a partial isometry, so $aa^*$ is also a projection and, being a nonzero projection in $qAq$, is equal to $q$. The identities $a^*a=p$ and $aa^*=q$ show that $x \mapsto ax : pAp \to qAp$ and $x \mapsto a^* x : qAp \to pAp$ are mutually inverse bijections, and the result follows.
Claim 5: Let $A$ be a unital $C^*$-algebra and let $p_1,\ldots,p_n$ be (necessarily orthogonal) projections satisfying $1_A=p_1 + \ldots + p_n$. Then, $A$ is the internal direct sum of the spaces $p_i A p_j$ where $1 \leq i,j \leq n$.
Basically, each element of $A$ can be thought of as a "matrix of operators" with respect to this partition of the identity.
Final Claim: Let $A$ be a $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then, $A$ is finite-dimensional.
Proof: First unitize $A$ if necessary (which does not affect the spectra of any of its elements). Then combine Claims 3, 4 and 5.
Best Answer
Look at $$\mathcal A= \{ f: [0,1]\to M_2(\Bbb C) \mid f(0) = a\,e_{11}, f(1)= b\,e_{22}, \ \ a,b\in \Bbb C\}$$ where $e_{ij}$ is the usual basis of $M_2(\Bbb C)$.
Any commutator must vanish at $0$ and $1$. Outside of those two points note that any matrix in $M_2(\Bbb C)$ is in the commutator ideal of $M_2(\Bbb C)$. In particular you may write $e_{ij} = \sum_k A_k [B_k, C_k]$, further you may write any function $f$ on $[0,1]\to\Bbb C$ as a product of $3$ functions to get:
$$f(x) e_{ij} = f_{ij}(x) \sum_k A_k[B_k,C_k]= \sum_k h^{1}(x) A_k [h^2(x) B_k, h^{3}(x) C_k]$$
where if $f$ vanishes on $0$ and $1$ you may assume all the $h^k$ to do so also. So:
Now lets check that $\Bbb C^2$ is not a sub-algebra of $\mathcal A$. The simplest way to do this is to note that $\mathcal A$ doesnt have two disjoint self-adjoint projections:
If $p$ is a projection then it must take on either the values $0$ or $e_{11}$ (resp $e_{22}$) on at $0$ (resp $1$). If we have two disjoint projections then at least one of them needs to take on $0$ at one of the endpoints (else the product is not zero). If a projection $p$ has $p(x)=0$ for some $x$ then there must be some $x'$ with $0<\|p(x')\|<1$ (else its constant $0$). But the equation $$p(x') \overset!= p(x')^2$$ cannot be satisfied for any self-adjoint matrix of norm greater than $0$ and less than $1$.