Just to flesh out Lucia's predicted negative answer a little bit using the older heuristic arguments from

<cite authors="Granville, Andrew">_Granville, Andrew_, [**Harald Cramér and the distribution of prime numbers**](https://doi.org/10.1080/03461238.1995.10413946), Scand. Actuarial J. 1995, No. 1, 12-28 (1995). [ZBL0833.01018](https://zbmath.org/?q=an:0833.01018).</cite>

(see also Prediction 17 of this [blog post of mine][1]).

Let $X$ be large. Let us restrict attention to intervals of the form $[WN-\log^2 X,WN+\log^2 X]$, where $W := \prod_{p \leq w} p$, $w := \frac{\log X}{\log\log X}$, and $X/2W \leq N \leq X/W$ is an integer.  We have $W = X^{o(1)}$, so the number of such intervals for a given $X$ is $X^{1-o(1)}$.  If each of these intervals has a "probability" of $\gg X^{-0.99}$ of containing fewer than $\log X$ primes, and we believe these "probabilities" to be "independent", then in analogy with the law of large numbers, we expect at least one of these intervals to furnish a counterexample to your inequality.

The point is that most of the numbers in the interval $[WN-\log^2 X,WN+\log^2 X]$ are already known to be composite: the only numbers that have a chance of being prime are of the form $WN \pm 1$, $WN \pm p$ for some prime $w < p \leq \log^2 X$, or $WN \pm pq$ for some primes $w < p,q$ with $p q \leq \log^2 X$.  One can check that the total number of candidates here is roughly $\frac{2\log^2 X}{\log(\log^2 X)} = \frac{\log^2 X}{\log\log X}$, by the prime number theorem (it is the $WN \pm p$ candidates that dominate).

On the other hand, the Cramer-Granville model (see previous reference) predicts that each such candidate has a "probability" of $\approx \frac{W}{\phi(W)} \frac{1}{\log X} \approx \frac{e^\gamma \log\log X}{\log X}$ of being prime (this calculation comes from Mertens' theorem).  If we believe these events to be "independent", then we expect the number of primes in $[WN - \log^2 X, WN + \log^2 X]$ to be distributed like a Poisson random variable of mean
$$ \lambda := \frac{\log^2 X}{\log\log X} \frac{e^\gamma \log\log X}{\log X} = e^\gamma \log X.$$

Now, a standard application of Stirling's formula shows that the probability that a Poisson variable of mean $\lambda$ is less than $\lambda (1+u)$ for some $-1 < u < 0$ is about $\exp(-\lambda h(u))$ where $h(u) := (1+u) \log (1+u)-u$, ignoring lower order terms (see e.g., [this blog post of mine][2], and compare also with [Bennett's inequality][3]).  Applying this with $u = e^{-\gamma}-1$, we predict that the "probability" of having fewer than $\log X$ primes is approximately
$$ \exp(-\lambda h(u)) \approx X^{-0.20386}$$
which is well above our target of $X^{-0.99}$, giving the desired prediction.  As noted by Lucia, the same analysis would in fact predict that one of these intervals would have $\lessapprox 0.27 \log X$ primes.

It's possible that a numerical search specifically targeting intervals of the above form (maybe after numerically optimizing the $w, X$ parameters to maximize the predicted probability of success for a given budget of search time, and maybe also shifting to $[WN, WN+2\log^2 X]$ instead which has marginally fewer candidates) may actually produce an explicit counterexample, although in practice convergence to these sorts of predictions is quite slow.

  [1]: https://terrytao.wordpress.com/2015/01/04/254a-supplement-4-probabilistic-models-and-heuristics-for-the-primes-optional/
  [2]: https://terrytao.wordpress.com/2022/12/13/an-improvement-to-bennetts-inequality-for-the-poisson-distribution/
  [3]: https://en.wikipedia.org/wiki/Bennett%27s_inequality