Let a function $f:\mathbb{R}\to \mathbb{R}$ be infinitely differentiable on $\mathbb{R}$. Suppose there exists $\delta>0$ such that
$0<x<\delta \implies f(a-x)<f(a)<f(a+x)$.
Then, does an interval $I\subset\mathbb{R}$ exist such that for all $x,y\in I$
$x<y \implies f(x)<f(y)$ ?
Note: not infinitely differentiable.
The function $f(x) = x(2+\sin(1/x)), x \ne 0$, and $f(0)=0$ is "increasing at $0$", but not increasing in any inverval around $0$.
You could similarly use $x^3(2+\sin(1/x))$ to make one that is differentiable everywhere.
(This assumes you intended to require $a \in I$)
Define $f:\mathbb R \rightarrow \mathbb R$ by
$$f(x)=\begin{cases} -e^{-1/x^2}\left[2 - \sin \left(\frac{1}{x}\right)\right] & \text{if} & x < 0\\ 0 & \text{if} & x=0\\ e^{-1/x^2}\left[2 - \sin \left(\frac{1}{x}\right)\right] & \text{if} & x < 0\end{cases}$$
Then $f$ is infinitely differentiable everywhere, and $f$ is strictly increasing at the point $x=0$ (in fact, $\delta = \infty$ in your requirement), and for each $\epsilon > 0$ we have $f$ is not monotone on the interval $(-\epsilon,\,0)$ and $f$ is not monotone on the interval $(0,\,\epsilon).$
To see that $f$ is strictly increasing at $x=0,$ note that $f(x)$ is negative when $x<0,$ $f(0)$ is zero, and $f(x)$ is positive when $x > 0.$
One way to prove $g^{(n)}(0)$ exists and equals $0$ for all positive integers $n$ is by mathematical induction similar to the method outlined here, the main difference for the function above being that one first proves by induction that each $g^{(n)}(x)$ has the form $e^{-1/x^2}\cdot\left[C_n(x)\cos(1/x) + S_n(x)\sin(1/x)\right]$ where $C_n$ and $S_n$ are rational functions of $x$ (whose explicit expressions are not needed).
Incidentally, I thought such an example could be found in Dini's 1878 book, given that his book has a similar example that I gave in my answer to Smooth function with infinite oscillation, but I wasn't able to find such an example in his 1878 book. Another useful old reference for real variable counterexamples is Esercizi e Note Critiche di Calcolo Infinitesimale by Ernesto Pascal (1895), but I didn't find an example there either.
If $f'(y)>0$ for some $y$ then $f$ is increasing around $y$. Now let $f'(y) \leq 0$ for all $y$. Then, for some $t$, $f(a+x)-f(a-x)=2x f'(t)\leq 0$ for all $x >0$. This contradicts the hypothesis.
If it holds for every $a$, then any interval of length $\delta$ will do.
If it holds for only one $a$, but that $a$ has to be in the interval, then how about this, whose derivatives at zero are all zero?
$$sign(x)[\exp(-x^{-4})+\exp(-x^{-2})\sin^2(1/x)]$$
