# Direct Proof that Continuous Functions Satisfy Epsilon-Delta Condition

If we define a continuous real function $$f: M \to R$$ as one that preserves sequential convergence: $$(x_n)\to p \Rightarrow f(x_n) \to f(p), \tag{1}$$ then to show that this definition is equivalent to the epsilon-delta condition, $$\forall p \forall \epsilon \exists \delta: x \in V_{\delta}(p) \Rightarrow f(x) \in V_{\epsilon}(f(p)), \tag{2}$$ my textbooks use use a proof by contradiction in which they derive a convergent sequence that does not converge under the mapping.

Question: Is it possible to have a direct proof of $$(1) \to (2)$$?

Thought: I think to directly show that $$(1) \to (2)$$, we will have to show that every element of M is an element of at least a sequence in $$M$$, i.e., the union of the elements of all the sequences in $$M$$ is $$M$$.