Please note that I am not asking for a proof, just a confirmation of my understanding or a counterexample to the question I posed.
We know that a function $f:M\to N$ for $M, N$ metric spaces is continuous if it preserves sequential convergence, i.e. if $(p_n)\to p$ in $M$, then $(f(p_n)) \to f(p)$ in $N$. In a problem I am currently working on, I am asked to show a weaker condition for continuity, which is stated as follows: if $f:M \to N$ is such that for every convergent sequence $(p_n)$ in $M$ we have that the sequence $(f(p_n))$ converges in $N$, then $f$ is continuous.
From my understanding, the convergence of the image sequence $(f(p_n)) \to q\in N$ where $q$ is not necessarily equal to $f(p)$ would contradict the definition of continuity where $f$ preserves sequential convergence. So how is this possible?
I don't know how to clear the doubt without illustrating what is pretty much a proof of what you seek. Let us use your "weak" condition prove "$p_n \to p \implies f(p_n) \to f(p)$".
So suppose your function $f:M \to N$ satisfies your weaker condition.
Now let $p \in M$ and let $(p_n)$ be a sequence in $M$ converging to $p$. Then from our assumption we know that $f(p_n) \to l \in N$. Now suppose $f(p) \neq l$. Then construct a sequence $(q_n)$ like,
$$ q_n = \begin{cases} p_n, & \text{if $n$ is even} \\ p, & \text{if $n$ is odd} \end{cases}$$
Then you will notice that $q_n \to p$ but the sequence $f(q_n)$ does not converge in $N$ contradicting our hypothesis.