Show that a meromorphic function on the complex plane, which achieves any complex number no more than fixed given times, must be rational.
If available, Picard's theorem is probably the quickest way to swat this fly [and it yields the result without assuming a fixed bound on the number of times a value is attained]. If Picard's theorem is not available or shall be avoided:
Let's call the meromorphic function $f$, and let's call the fixed bound on the number of times any complex number is attained $n$.
If $f$ has at least $k$ poles, then every complex number of sufficiently large modulus is attained at least $k$ times, so $k \leqslant n$. Let $\{\pi_1,\dotsc, \pi_p\}$ be the set of poles of $f$.
Let $m$ be the maximum number any complex value is attained at all (counting multiplicities). Then $m \leqslant n$. Pick $w_0 \in \mathbb{C}$ such that $w_0$ is attained $m$ times (counting multiplicities), and choose $R$ so large that $f(z) \neq w_0$ for $\lvert z\rvert \geqslant R$, and that $\lvert \pi_k\rvert < R$ for all poles of $f$. Choose $\varepsilon > 0$ small enough that $0 < \lvert z - \pi_k\rvert \leqslant \varepsilon \implies \lvert f(z)\rvert \geqslant 1 + \lvert w_0\rvert$ and $\lvert \pi_k\rvert + \varepsilon < R$. Let $G = \{ z\in \mathbb{C} : \lvert z\rvert < R, \lvert z-\pi_k\rvert > \varepsilon\}$. Then look at
$$N(w) := \frac{1}{2\pi i} \int_{\partial G} \frac{f'(z)}{f(z) - w_0}\,dz.$$
By construction, $N(w_0) = m$, and $N$ is constant on each component of $\mathbb{C}\setminus f(\partial G)$, so we have $N(w) \equiv m$ on some neighbourhood $U$ of $w_0$, which means that every $w\in U$ is attained $m$ times (counting multiplicities) in $G$. By the maximality of $m$, that means no $w\in U$ is attained outside $G$, in particular
$$U \cap f\bigl(\mathbb{C}\setminus \overline{D_R(0)}\bigr) = \varnothing.$$
By the Casorati-Weierstraß theorem, it follows that $\infty$ is not an essential singularity of $f$. Then the rationality of $f$ follows by subtracting the principal parts of $f$ at the $\pi_k$ and (possibly) at $\infty$.
Best Answer
The exponential function has an essential singularity at $\infty$, but is meromorphic everywhere else -- and is not rational.
On the other hand, if the singularity at $\infty$ is removable, then the proof you sketch still works, and shows that $f$ must be a rational function.