Showing $f$ is meromorphic or that $f(a_j)$ converges to some point in the plane. 
Let $S$ be a sequence of points in $\mathbb{C}$ that converges to $0$. Let $f$ be analytic and defined on some disc centered at $0$ except possibly at the points of $S$ and at $0$. Show that either $f(z)$ extends to be meromorphic in some disc containing $0$, or else for any complex number $w$ there is a sequence $a_j$ such that $a_j\to 0$ and $f(a_j)\to w$ as $j\to \infty $.

Suppose $f$ cannot be extended to a meromorphic function in $D(0,r)$ for some $r>0.$
Why does it suffice to consider the two cases:
Case 1: There is a sequence $z_j$ in $S$ such that $z_j$ is an essential singularity of $f$. (So by Big Picard, we can find $a_j$ such that $f(a_j)\to w$.)
Case 2: There is a $D(0,r)$ for which any $z_n\in D(0,r)$  is either a pole or a removable singularity (we can ignore the later since in that case we can extend the function to the singularity by Riemann).
I get why we consider case 1 since meromorphic functions don't have essential singularities. Why do we need case 2?
 A: We need case 2, since if $f$ is meromorphic on $D_r(0) \setminus \{0\}$ and has a sequence of poles converging to $0$, then $0$ is not an isolated singularity of $f$, and thus Picard's theorem doesn't apply. It may be that Picard's theorem applies to a simple transformation of $f$, consider e.g.
$$f(z) = \frac{1}{\sin \frac{1}{z}},$$
where $1/f$ has an isolated essential singularity at $0$, but if
$$f(D_\rho(0)\setminus \{0\}) = \widehat{\mathbb{C}}$$
for all $0 < \rho \leqslant r$, there is no easy transformation of $f$ making Picard's theorem applicable.
Rather than invoking big guns like Picard's theorem, I would prefer a more elementary argument, however.
If there is a $w\in \mathbb{C}$ such that there is no sequence $(a_j)_{j\in\mathbb{N}}$ in $D\setminus (S\cup \{0\})$ with $a_j \to 0$ and $f(a_j) \to w$, then there are $r > 0$ and $\varepsilon > 0$ such that $\lvert f(z) - w\rvert \geqslant \varepsilon$ for all $z\in D_r(0) \setminus (S\cup \{0\})$, and then
$$g(z) = \frac{1}{f(z)-w}$$
is a bounded holomorphic function on $D_r(0) \setminus (S\cup \{0\})$. Applying the removable singularity theorem to the isolated singularities of $g$ in $D_r(0)\cap S\setminus \{0\}$, we obtain a holomorphic extension of $g$ - which we still denote by $g$ - to $D_r(0)\setminus \{0\}$, and now we can apply the removable singularity theorem to the isolated singularity $0$, yielding a holomorphic extension $h$ of $g$ to $D_r(0)$. But then
$$\tilde{f}(z) = w + \frac{1}{h(z)}$$
is a meromorphic extension of $f$ to $D_r(0)$.
