Quotients of finite C-star algebras: are they finite? Edit: to whomever downvoted this post: please explain why. Seriously, I'd like to know why because I cannot think of any reason. This is a fine question, it got 3 upvotes, in my humble opinion it is interesting and it deserves to be here. There are also two beautiful answers by Ruy and PStheman presented here to address this question. So why the downvote?
Edit 2: Now I see someone retracted their upvote. Please explain why you changed your mind. What is it that you saw and decided that, on second thought, this post does not deserve your encouragement and support? I would really like to see some explanations for all this drama.
Edit 3: Another downvote. Makes one wonder, why is this happening?

Original post
As I was trying to answer this question, I thought of the following: If $A$ is a unital, finite $C^*$-algebra and $I$ is an ideal in $A$, then is $A/I$ a finite $C^*$-algebra?
Recall that a unital $C^*$-algebra is called finite when the implication
$$p\sim_{M-vN}1_A\implies p=1_A$$
holds, i.e. every isometry in $A$ is a unitary. Besides the main question, I realized something: I do not know what $C^*$-algebras are finite or not. For example, I know that abelian $C^*$-algebras are stably finite, as it is (briefly) discussed in this post, and finite dimensional $C^*$-algebras are (stably) finite too, but that's all I know. So a bonus question is: what are some interesting examples of finite $C^*$-algebras besides abelian and finite dimensional ones?
Note that in the answer of the linked question I am using a particular feature of the ideal $\sum_nA_n$ in $\prod_nA_n$, something that we cannot do in the general case.
 A: Consider
$$
  A = \{f\in C([0, 1], \mathcal T): f(0)\in  \mathbb C\cdot 1\},
  $$
where  $\mathcal T$ is the Toeplitz algebra.  Every isometry $f$ in $A$ gives rise to a path of isometries in
$\mathcal T$ beginning with a scalar.  Observing that  all isometries in the Toeplitz algebra have finite Fredholm
index, and also that the index is continuous, we can easily show that $A$ is finite.  However
the map
$$
  f\in  A\mapsto  f(1)\in \mathcal T
  $$
is a surjection from $A$ to a non-finite algebra.

EDIT. Here is an argument to show the finiteness of $A$ which does not rely on the theory of  Fredholm operators:
Lemma.  Let $B$ be a unital C*-algebra.  Denote by   $\mathscr I(B)$  the set of all isometries in $B$, and by   $\mathscr U(B)$ the subset
formed by the unitary elements.  Then $\mathscr U(B)$ is clopen in $\mathscr I(B)$.
Proof.  It is well known that the set  $\text{GL}(B)$ formed by all invertible elements is open.  Since
$$
  \mathscr U(B) = \mathscr I(B) \cap \text{GL}(B),
  $$
it follows  that   $\mathscr U(B)$ is open in $\mathscr I(B)$.  To show that $\mathscr U(B)$ is also closed,  suppose by
contradiction that the
sequence $\{u_n\}_n$  of unitaries converges to a proper isometry $s$ (meaning that $ss^*\neq 1$).
The element
$$
  p = 1-ss^*
  $$
is then a non-trivial projection and,  clearly, $ps=0$.  We then deduce that
$$
  0 = ps = \lim_{n\to \infty }pu_n,
  $$
but this is a contradiction since $\|pu_n\|=\|p\| = 1$.  QED
This said,  suppose that $f$ is an isometry in $A$,  so we have that   $f(t)$ lies in $\mathscr I(\mathcal T)$, for all
$t$ in $[0, 1]$.
Since $f(0)$ is necessary a scalar, we have that $f(0)\in \mathscr U(\mathcal T)$, but then
$f(t)\in \mathscr U(\mathcal T)$, for every $t$ because  $[0,1]$ is connected.
This shows that $f$ is unitary and hence that
$A$ is finite.
A: There is probably an obvious example laying around, but I can't think of one! Here is an overkill answer showing that quotients of finite C*-algebras need not be finite.
There is a characterization of separable exact C*-algebras as subquotients (quotient of a subalgebra) of the CAR algebra, i.e., the UHF algebra $M_{2^{\infty}}$. This algebra is finite and so is any subalgebra since the faithful trace on the CAR algebra restricts to a faithful trace on any smaller algebra. So take $B \subseteq M_{2^{\infty}}$ ($B$ unital) and $I \lhd B$ such that $B/I \simeq \mathcal{O}_2$, the Cuntz algebra generated by 2 isometries. The Cuntz algebra is (purely) infinite.
There is a nice description of finiteness in unital C*-algebras (see Left Invertibility Implies Right Invertibility in Certain $C^*$-algebras). Some examples are UHF algebras (like the CAR algebra), the irrational rotation algebra, the Jiang-Su algebra, etc.
References:
This was originally proved in "On subalgebras of the CAR-algebra" by Kirchberg, and later Wassermann gave a "simpler" (take with a grain of salt) proof in "Subquotients of UHF C*-algebras".
