Proving a function from metric space to another is continuous? I was practicing some problems from my Analysis text last night and I came across this one. 
I'm a bit stumped as to how to approach it... 
Let's assume that f: M -> N is a function from one metric space to another which follows the condition that if a sequence (pn) in M converges then the sequence (f(pn)) in N converges. 
How would you go about proving that f is continuous? I feel as if though I'm overthinking it, but my work isn't aligning properly. 
 A: Suppose that $f$ is not continuous at $a\in M$. Then there is a sequence $\{x_n\}\subset M$  that converges to $a$ but $f(x_n)$ does not converge to $f(a)$. The sequence
$$
x_1,a,x_2,a,x_3,a,\dots,x_n,a,\dots
$$
also converges to $a$, and the sequence
$$
f(x_1),f(a),f(x_2),\dots,f(x_n),f(a),\dots
$$
is convergent. Can you arrive at contradiction from here?
A: Assume that the function is not continuous and that for every convergent sequence in $M$, the image is convergent in $N$. Since $f$ is not continuous, there exists a point $y:=f(x)\in f(N)$ and an $\epsilon>0$, such that for $V:=B(y;\epsilon)$ (the open ball of radius $\epsilon$ around $y$), we have that for every $\delta>0$, we find $x'\in B(x;\delta)$ such that $f(x')\notin V$. Now let $x_i\in B(x;\frac{1}{i})$ satisfy this condition. Then the sequence $x_n$ clearly converges to $x$, but $f(x_n)$ does not converge to $y$ (since it always stays at least $\epsilon$ far away). Now consider the sequence $x_1,x,x_2,x\cdots$. This converges to $x$, hence $f(x_1),f(x),f(x_2),f(x)$ converges. Since limits in metric spaces are unique, this sequence must converge to $y$ (since $f(x),f(x),\cdots$ does). Now we arrive at the contradiction that $f(x_n)$ does converge to $y$, which previously was stated impossible (thanks to Julián Aguirre for the last trick).
A: Hint: $f$ is continuous $\Leftrightarrow$ For every subset $A$ of $M$, $f(cl(A)) \subset
cl(f(A))$, where $cl()$ is the closure.
Work with closure is work with limits points
