Let $f$ be a function from a metric space to a metric space. Prove that $f$ is continuous i.f.f it sends convergent sequences into convergent seq. My question is as stated in the title; Let $f$ be a function from a metric space to a [second] metric space. Prove that $f$ is continuous if and only if it sends convergent sequences into convergent sequences.
To be honest, I am a bit stuck on this one and would appreciate any input how to get started. What is the intuition? 
 A: Hint:
Let $\{a_n\}$ be any convergent sequence in $X$ and suppose that $\lim_{n\to\infty}a_n=a$, i.e. $\forall \delta>0$, $\exists M > 0$ such that
$$n\geq M \implies |a_n-a| < \delta$$
Now because $f: X \to Y$ is continuous at $x=a$, for any $\epsilon>0$ you will find a positive $\delta>0$ such that $|x-a|<\delta \implies |f(x)-f(a)| < \epsilon$. How can you combine these two facts?
To prove the other direction, suppose that $f$ is not continuous at some point $x=a$, then $\exists\epsilon_{\star}>0$ such that for all $\delta>0$, we have $|x-a|<\delta$ but $|f(x)-f(a)| \geq \epsilon_{\star}$. Take $\delta = \frac{1}{n}$ to construct a sequence $\{a_n\}$ such that $a_n\to a$. 
Now define a new sequence $\{b_n\}$ such that
$$b_n=\begin{cases}
a_n & \text{if $n$ is odd} \\ a & \text{if $n$ is even}
\end{cases}$$
Then $\lim_{n\to\infty}b_n=a$ still remains true but $f(b_n)$ is NOT a Cauchy sequence because $|f(b_{2n+1})-f(b_{2n})|=|f(a_{2n+1})-f(a)| \geq \epsilon_{\star}$. Because any convergent sequence is a Cauchy sequence, the sequence $\{f(b_n)\}_{n=1}^{\infty}$ does not converge. Contradiction.
