Theorem: If $f$ is continuous on the interval $[a,b]$ then $\forall y \in [f(a),f(b)], \space \exists c \in [a,b]$ such that $f(c)=y$
I've seen many proofs using the epsilon-delta definition of continuous maps. So I wanted to prove it using a slightly different approach using the definition of continuous maps as maps that preserve sequential convergence.
Proof attempt:
Define $$A:=\{x \in [a,b] : f(x)<y\}$$
Claim: if $c=\sup A$, then $f(c)=y$
We show it by eliminating other possibilities. Suppose $f(c)<y$. Then let $(x_{n})$ be a sequence with $(x_{n}) \to c$ and $x_{n} >c$ for all $n \in \mathbb{N}$. The continuity of $f$ preserves the sequenial convergence of $(x_{n})$ so $f(x_{n}) \to f(c)$. If $(f(x_{n_{k}}))$ is a monotonic subsequence of $(f(x_{n}))$, then we know that there exists $N \in \mathbb{N}$ such that $\forall n^*\geq N$ where $n^* \in \{n_{k}\}$, we have $f(c)<f(x_{n^*})<y$. But then $x_{n^*} \in A$ but $c= \sup A<x_{n}$ for all $n \in \mathbb{N}$. So $f(c)\geq y$.
On the other hand, if $f(c)>y$, we can similarly construct a sequence with $(x_{n}) \to c$ but $x_{n} \in A$ for all $n \in \mathbb{N}$. We again find $N \in \mathbb{N}$ such that $\forall n^*\geq N$ we have $y<f(x_{n^*})<f(c)$. But then $x_{n^*}$ is an upper bound which impossible since $x_{n}<c$ for all $n \in \mathbb{N}$.
We can conclude $f(c)=y$