Setup:
Suppose $M$ is a $C^k$-manifold embedded into some Hilbert space $H$ and $g$ is the induced Riemannian metric thereon (induced by restricting the inner-product $\langle,\rangle_H $ in $H$ to the tangent spaces of $M$).
Let $g_H(x,y):= \sqrt{\langle (x-y),(x-y)\rangle_H}$.
Question:
Then is it necessarily the case that: \begin{align} (\forall x,y \in M)\, g_H(x,y)\leq g(x,y)? \end{align}
Intuitive Reasoning
This seems to intuitively be true as the "shortest distance" between two points is a straight line and $M$ may be curved..
To make your intuition precise, we should prove that the arclength $\int |\dot \gamma|$ of any ($C^1$) curve $\gamma$ joining $x$ and $y$ is at least $|x-y| = g_H(x,y)$.
Let's start by defining a "horizontal axis" to measure along: Let $v = \frac{x-y}{|x-y|}$ and define $h(p) = \langle p-y, v \rangle$, so that $h$ is an affine function with $h(y) = 0$, $h(x) = |x-y|$ and unit-length gradient (don't worry about making this last claim precise).
We want to show that the arclength is bounded below by $h(\gamma(1))$ and we know that $h(\gamma(0))=0$, so it suffices to show that
$$\frac{d}{dt} h(\gamma(t)) \le |\dot \gamma| = \frac{d}{dt} \int_0^t |\dot \gamma|.$$
This is now simple - just apply Cauchy-Schwarz:
$$ \frac d{dt} h(\gamma(t)) = \frac d{dt} \langle \gamma(t) - y, v\rangle = \langle \dot \gamma(t),v \rangle \le |\dot \gamma||v| = |\dot \gamma|.$$