It's true that $$\frac{1}{|B_r(x)|}\int_{B_r(x)} |u(y)-u(x)|^p\to 0$$ for $r\to 0$. For $p=1$ this is standard (Somehow I always overlooked the absolute value signs in the definition of Lebesgue points), and the same proof carries over for $p>1$ (at least the proof used in Ziemer, "Weakly differentiable functions", where variants of this theorem for Sobolev spaces are discussed as well).
Every locally compact metric space can be given a compatible complete metric.
Suppose that $X$ is a locally compact metric space. Then $X$ is paracompact, so $X$ is a disjoint union of $\sigma$-compact locally compact spaces (There is a theorem proved in the book Topology by Dugundji that proves that every paracompact locally compact space is the disjoint union of $\sigma$-compact spaces. I prove this fact below.). Therefore we may without loss of generality assume that $X$ is $\sigma$-compact. Since we assume $X$ is $\sigma$-compact, there is a sequence of open sets $U_{n}$ such that $\overline{U_{n}}$ is compact and $\overline{U_{n}}\subseteq U_{n+1}$ for all $n$. From this sequence of open sets and Urysohn's lemma, there is a function $f:X\rightarrow[0,\infty)$ such that $\overline{U_{n}}\subseteq f^{-1}[0,n)\subseteq U_{n+1}$. Let $d$ be a metric on $X$, and define a new metric $d'$ on $X$ by letting $d'(x,y)=d(x,y)+|f(x)-f(y)|$. Clearly $(X,d')$ induces the original topology on $X$. I claim that $(X,d')$ is complete. Assume that $(x_{n})_{n}$ is a Cauchy sequence in $(X,d')$. Then the sequence $(x_{n})_{n}$ is bounded in $(X,d')$, so clearly the sequence $(f(x_{n}))_{n}$ is bounded as well. Therefore, there is some $N$ where $f(x_{n})<N$ for all $n$. In particular, since $f(x_{n})\subseteq f^{-1}[0,N)$, we have $x_{n}\in U_{N+1}\subseteq\overline{U_{N+1}}$ for all $n$. Since $\overline{U_{N+1}}$ is compact, and $x_{n}\in\overline{U_{N+1}}$ for all $n$, the sequence $(x_{n})_{n}$ has a convergent subsequence, so the sequence $(x_{n})_{n}$ itself must be convergent.
Added Later I will now prove that a paracompact locally compact space is a free union of $\sigma$-compact spaces since the proof is not too difficult.
Suppose that $X$ is locally compact and paracompact. Then let $\mathcal{U}$ be a cover of $X$ such that if $U\in\mathcal{U}$, then $\overline{U}$ is compact. Then let $\mathcal{V}$ be a locally finite open refinement of $\mathcal{U}$.
I claim that each $U\in\mathcal{V}$ intersects only finitely many other elements in $\mathcal{V}$. Since $\mathcal{V}$ is locally finite, there is an open cover $\mathcal{O}$ of $X$ where each $O\in\mathcal{O}$ intersects only finitely many elements of $\mathcal{V}$. If $U\in\mathcal{V}$, then since $\overline{U}$ is compact, there are $O_{1},...,O_{n}\in\mathcal{O}$ where $U\subseteq\overline{U}\subseteq O_{1}\cup...\cup O_{n}$. However, since the sets $O_{1},...,O_{n}$ each only intersect finitely many elements of $\mathcal{V}$, the union $O_{1}\cup...\cup O_{n}$ can only intersect finitely many elements of $\mathcal{V}$. In particular, the set $U$ only intersects finitely many elements of $\mathcal{V}$. Now make the set $\mathcal{V}$ into a graph where we put an edge between $U,V\in\mathcal{V}$ if and only if $U\cap V\neq\emptyset$. Then each $U\in\mathcal{V}$ has only finitely edges. Let $(\mathcal{E}_{i})_{i\in I}$ be the set of all connected components of the graph $\mathcal{V}$. Then each $\mathcal{E}_{i}$ is a countable subset and clearly $\{\bigcup\mathcal{E}_{i}|i\in I\}$ partitions $X$ into clopen subsets. Therefore since each $\mathcal{E}_{i}$ is countable and each $U\in\mathcal{V}$ has compact closure, the each union $\bigcup\mathcal{E}_{i}$ is $\sigma$-compact.
Best Answer
I assume by Lebesgue differentiation theorem you mean the statement that $|B(x)|^{-1}\int_{B(x)} f(y)\, dy \to f(x)$.
Then it's not clear to me what your set-up on $X=[0,1]^{\mathbb N}$ is (what's the measure?), but in any event, for an arbitrary metric, this already fails on $\mathbb R^2$. You can take a metric that gives you wide thin rectangles as small balls, for example $$ d(x,y)=\max (|x_2-x_1|, |y_2^{1/3}-y_1^{1/3}|) $$ (if $y<0$, then $y^{1/3}$ just means $-|y|^{1/3}$).
Update: This answer was originally based on my recollection of the "standard fact" that the higher-dimensional Lebesgue differentiation theorem fails for rectangles if the eccentricity is not restricted. This much is true if arbitrary rectangles are allowed, but of course the situation here is different, and I'm not sure now what the situation is (and in fact I'm not even sure it's not an open question).