Let $(X, d)$ and $(X^\prime, d^\prime)$ be metric spaces and $f: (X, d) \to (X^\prime, d^\prime)$ be a function. Consider the following (equivalent) definitions of continuity at $x \in X$:
(P1): $\forall \epsilon > 0, \exists \delta > 0 \hspace{0.5em} \text{s.t.} \hspace{0.5em} \forall x^\prime \in X, d(x^\prime, x) < \delta \implies d^\prime(f(x^\prime), f(x)) < \epsilon$
(P2): $\forall \{x_n\}_{n=1}^\infty (\subseteq X), \ x_n \to x \implies f(x_n) \to f(x)$
The proof of $\text{(P1)} \implies \text{(P2)}$ is easy and straightforward. The proof of $\text{(P2)} \implies \text{(P1)}$ is also easy if we show $\lnot \text{(P1)} \implies \lnot \text{(P2)}$.
However I do not come up with nor see the direct (not using the contrapositive relation) proof $\text{(P2)} \implies \text{(P1)}$. Does anyone know it?
P.S.
@Tom Collinge: Thnak you for your comment. I need to clarify my question: Is there a way to prove $\text{(P2)} \implies \text{(P1)}$ without assuming $\lnot \text{(P1)}$?. Although proof by contrapositive and proof by contradiction are not equivalent in general, I think steps of the two proofs are almost same in this question.
@mechanodroid and @LoveTooNap29: Thank you for helpful comments! I saw Brian M. Scott's answer for this question and still have following unclear points
(minor) when we prove $\text{[sequentially continuous]} \implies \text{[$\epsilon$-$\delta$ continuous]}$, we need the axiom of countable choice (CC), whether using proof by contrapositive or not (contrapositive case is mentioned in the question).
(major) I think Brian M. Scott's answer uses a kind of $\lnot \text{(P1)} \implies \lnot \text{(P2)}$ discussion in the proof of his Claim:
If $f$ is sequentially continuous at $x$ (P2), then $f \upharpoonright (\mathbb{Q} \cup \{x\})$ is $\epsilon$-$\delta$ continuous at $x$ (P1)
