**THM** Let $(M,d)$ be a complete metric space with no isolated points. Then $(M,d)$ is uncountable.

**PROOF** Assume $M$ is **countable**, and let $\{x_1,\dots\}$ be an enumeration of $M$. Since each singleton is closed, each $X_i=X\smallsetminus \{x_i\}$ is open for each $i$. Moreover, each of them is dense, since each point is an accumulation point of $X$. By Baire's Theorem, $\bigcap_{i=1}^n X_i$ must be dense, hence nonempty, but it is readily seen it is empty, which is absurd.

**COROLLARY** Let $(M,d)$ be complete, $P$ a perfect subset of $M$. Then $P$ is uncountable.

I haven't taken the putnam yet.
However, I am preparing for the Putnam 2014. My strategy is to prepare intensely over a long period of time. Note that I have no experience in Math Olympiad.

As a college student, I have access to my college's Putnam resources. For example, here at UCLA there is a Putnam coach (Prof. Ciprian Monalescu!)and a Putnam preparation program.

Your concern about classes+work+time is very real. There is an entire thread on CollegeConfidential, http://talk.collegeconfidential.com/science-majors/1306946-chances-putnam-exam.html .