This question could be seen by some as too basic, but I think it is precisely the kind of situation MO is for (question is not trivial to an outsider, but easy to answer to an insider of the field), so let me give a complete argument to show that the property you seek is true for any metric (only adding some flesh to the comments).

**Edit** *Here is a modified argument, applying to continuous metrics (and not resorting to the exponential map) even when one wants to control the induced distance on the neighborhood, rather than its length metric.*

Let $g$ be any continuous Riemannian metric on $M$ and fix $p\in M$. Consider any chart $\varphi: B\to M$ where $B$ is the unit Euclidean ball, such that
$\varphi(0)=p$. Compare the metrics $g$ and $\tilde g:=\varphi_* g_0$ (where $g_0$ is the Euclidean metric) on the image $D:=\varphi(\frac12 B)$ of the ball of radius $1/2$. Since both are continuous (here we need the chart to be $C^1$, but not more), there are $k,K$ such that $k\tilde g\le g \le K\tilde g$ on the closure of $D$.

Denote by $\ell(\gamma)$ the length of a curve $\gamma$ on $D$ with respect to the metric $g$, and $\tilde\ell(\gamma)$ its length with respect to the metric $\tilde g$. Then the formula for length of curves immediately gives
$$ k\tilde\ell(\gamma) \le \ell(\gamma)\le K\tilde\ell(\gamma).$$
By taking the infimum, we obtain that the length metrics induced by $g$ and $\tilde g$ on $D$ are bi-Lipschitz.

Now, there is a small enough $r>0$ such that $k(1-2r)\ge 2K r$; consider the smaller Euclidean ball $rB$ and its image $D_r:=\varphi(rB)\subset D$. I claim that the restriction to $D_r$ of the length metric of $(M,g)$ is bi-Lipschitz to the length metric induced on $D_r$ by $\tilde g$.

The only thing to check is that we can restrict to curves on $D$, i.e. that given $x,y\in D_r$, there is no curve $\gamma$ on $M$ between them which is shorter than $\ell(\tilde\gamma)$, where $\tilde\gamma$ is the $\tilde g$ geodesic from $x$ to $y$ (which is the $\varphi$-image of a Euclidean segment). If there where, the curve could not stay in $D$, so that $\gamma$ would cross twice the annulus $D\setminus D_r$. Its $\tilde g$ length would be at least $1-2r$, hence its $g$ length would be at least
$$k(1-2r)\ge 2Kr\ge K\tilde\ell(\tilde\gamma)\ge \ell(\tilde \gamma)$$
and we are done.

Of course, by compactness of $M$, one can take the constants $k$ and $K$ uniform over $M$. For a non-compact $M$, the above still holds non-uniformly.

*The previous argument*

Cover your manifold with finitely many charts $(\varphi_i)$ defined on Euclidean balls $(B_i)$, such that the images of the balls with half radius $(\frac12 B_i)$ still cover $M$. Each point has $D_i:=\varphi(\frac12 B_i)$ as a neighborhood. Compare for that chart $\varphi_i$ the metric $g$ on $\varphi_i(B_i)$ and the pushforward of the Euclidean metric, $\tilde g=\varphi_*g_0$. This second metric is *isometric* to a Euclidean ball by definition.

Since both metrics are smooth (in fact we only need them to be continuous, as mentioned by Deane Yang), there are constants $k$ and $K$ such that $k\tilde g\le g \le K\tilde g$ on the closure of $D_i$ (here is why we took balls of half radius, to use compactness).

Denote by $\ell(\gamma)$ the length of a curve $\gamma$ on $D_i$ with respect to the metric $g$, and $\tilde\ell(\gamma)$ its length with respect to the metric $\tilde g$. Then the formula for length of curves immediately gives
$$ k\tilde\ell(\gamma) \le \ell(\gamma)\le K\tilde\ell(\gamma)$$
and passing to the infimum, you get that $\varphi$ is bi-Lipschitz (in fact, all of this is a lengthy spelling-out of the fact that diffeomorphisms between Riemannian manifolds are always locally Lipschitz!)

Now, if you want to compare with a Euclidean metric the restriction to $D_i$ of the length metric on $M$ (as opposed to the length metric of the restriction of $g$ to $D_i$), then there is a small issue left: the geodesic of $M$ between two points of $D_i$ might go outside $D_i$. Then you can use the exponential map instead of any chart, which has the additional property that image of small balls are convex, taking care of this point.

everypoint ineveryRiemannian manifoldeverysmall enough neighborhood is bi-Lip to the unit ball in the Euclidean space seems quite basic to me. $\endgroup$non-compactanduniformbi-Lipschitz constants are discussed. In any case, I am not an expert, so maybe I am missing something obvious, moreover the MO system starts shouting that I am being chatty, so I'll stop it here. $\endgroup$any(finite) cover of $M$ of balls. Since $g$ is positive definite on the closure of each ball, there exists a positive constant $a > 0$ such that $a^{-1} \le g \le a$ and therefore the ball with the Riemannian metric structure is bilipschitz to the Euclidean ball. $\endgroup$7more comments