Let $x: U \subset \mathbb{R^2} \to\mathbb{R^3}$ be a regular parametrized surface such that $x$ is injective and let $S = X(U)$. Let $p_1, p_2 \in S$ be fixed and consider the family of regular curves $\beta \subset S$ that pass through $p_1$ and $p_2$. Let $\alpha \in \beta$ such that $\alpha(t_1) = p_1$ and $\alpha(t_2) = p_2$. The intrinsic distance from $p_1$ to $p_2$ on $S$ is denoted by: $$d(p_1, p_2) = \inf\left(\int_{t_1}^{t_2} \|\alpha'(t)\| \ dt \right)$$
Prove that:
a) $d(p_1, p_2) \geqq |p_1-p_2|$
b) Let $X$ and $\overline{X}:U \to \mathbb{R^3}$ be isometric surfaces such that $S = X(U)$ and $\overline{S} = \overline{X}(U)$. Prove that the isometry $\phi: S \to \overline{S}$ preserves the intrinsic distance between corresponding points, i.e $p_1, p_2 \in S \implies d(p_1, p_2) = \overline{d}(\phi(p_1),\phi(p_2))$, where $d$ and $\overline{d}$ are the intrinsic distances of $S$ and $\overline{S}$, respectively.
I have thought a lot on this and can't find an approach to a) or b). Any help/hints on useful starting points/ would be very appreciated.
Assume for simplicity that $p_1=(0,0,0)$ and $p_2=(1,0,0)$ while $t_1=0$ and $t_2=1$, any other constellation can be obtained by reparametrizing the curve and moving the coordinate frame. Now take any curve $\alpha = (\alpha_1,\alpha_2,\alpha_3)$ connecting $p_1$ and $p_2$, then $$ \int_0^1\Vert\alpha'(t)\Vert dt \ge \int_0^1\vert\alpha'_1(t)\vert dt \ge \int_0^1\alpha'_1(t)dt=\alpha_1(1)-\alpha_1(0) = 1 = \Vert p_1 - p_2 \Vert. $$