I think there isn't. Here's a sketch of a proof. I'm just not sure whether it really works because I'm not confident with the transfinite versions of the standard theorems about limits and convergent sequences.
Attempted proof: Suppose there were such a function. Then use transfinite recursion to generate the following transfinite sequence:
BASE CASE: $s_0 = 1$.
SUCCESSOR ORDINAL: Let $s_{\lambda + 1}$ be a number in $[0, 1]$ such that $f(s_{\lambda + 1}) < f(s_\lambda)$.
LIMIT ORDINAL: Suppose $\lambda$ is a limit ordinal. Then the sequence $f(s_0)$, ..., $f(s_\alpha)$, ... with $\alpha < \lambda$ is bounded below by 0. So it has a limit $\lim_{\alpha < \lambda} f(s_\alpha$) (by a transfinite version of the Monotone Convergence Theorem). Also (by a transfinite version of the Bolzano-Weierstrass Theorem) there is a subsequence of $s_0$, ..., $s_\alpha$, ... that is unbounded in that sequence and it converges to a limit. Let's call that subsequence $s_{i_0}$, ..., $s_{i_\beta}, ...$ for $\beta < \gamma \leq \lambda$. Then let $s := \lim_{\beta < \gamma} s_{i_\beta}$ Then, since $f$ is continuous, $$ \lim_{\alpha < \lambda} f(s_\alpha) = \lim_{\beta < \gamma} f(s_{i_\beta}) = f(s) $$ So let $s_\lambda = s = \lim_{\beta < \gamma} s_{i_\beta}$.
Then there must be $\alpha$ such that $s_\alpha = s_{\alpha + 1}$, since there are at most continuum-many distinct numbers in the list $f(s_0)$, ..., $f(s_\alpha)$, ... . This gives a contradiction.
There is a rather simple proof using basic analysis: A continuous function $f$ on the compact interval $[0,1]$ is always bounded and attains its minimum (Extreme value theorem). If $x \in [0,1]$ is such that $f(x) = \inf_{[0,1]} f$, then there would be no $y$ with $f(y) < f(x)$.
Proof of the existence of a minimum: Let $M=\inf_{[0,1]} f$. Then there is a sequence $(x_n)_n$ such that $f(x_n) \to M$. Since $[0,1]$ is compact there is a convergent subsequence $(x_{n_k})_k$. You can show this using the Bolzano-Weierstraß-Theorem. Let $x = \lim_k x_{n_k}$, then $f(x) = \lim_k f(x_{n_k}) = \lim_n f(x_n) = M$.
I don't know about your limit ordinal proof, but using transfinite methods and thereby the axiom of choice is not necessary for this proof.