Homotopy problem Let $X$ and $Y$ metric space with $X$ compact and $f,g: X \rightarrow Y$ homotopic. Then for each $\epsilon>0$, exists continuous functions $f_0,\ldots,f_k: X \rightarrow Y$ such that:
$$ d(f_i(x),f_{i-1}(x))<\epsilon, \quad \forall x\in X $$
and $f_0=f$, $f_k=g$.
I was thinking of using the fact that $H: X \times [0,1] \rightarrow Y$ is continuous in a compact, so is uniformly continuous. But I can't prove what it asks for
 A: The homotopy $H$ tells you how to move continuously from $f$ to $g$, and the problem asks you to move from $f$ to $g$ in small, discrete steps. Getting discrete steps is just a matter of discretizing the continuous motion that you already have. Subdivide the parameter interval $[0,1]$ into little pieces, say $[\frac{i-1}N,\frac iN]$ for some big $N$ and $i=1,2,\dots,N$. Then take snapshots of the motion $H$ at those subdivision points; that is, set $f_i(x)=H(x,\frac iN)$. (Another rough description of this: Watch the motion $H$ with a stroboscope.) That gets you from $f_0=f$ to $f_N=g$ in discrete steps.
But the steps are also required to be small. To achieve that, you want to take $N$ big enough so that the intervals $[\frac{i-1}N,\frac iN]$ are very short. Here, your idea of using the uniform continuity of $H$ solves the problem. For any desired $\varepsilon$ in the problem, find a $\delta$ as in the definition of uniform continuity, and take $N>\frac1\delta$.
A: You can use the Lebesgue number lemma.
Let $\cal V$ be the open cover of $Y$ by open balls of radius $\epsilon/2$ around each point of $Y$.
Let $\mathcal U = \{H^{-1}(V) \mid V \in \mathcal V\}$, which is an open cover of the compact space $X \times [0,1]$.
Applying the Lebesgue number lemma, and using the ordinary product metric on $X \times [0,1]$, let $\lambda > 0$ be a Lebesgue number for $\mathcal U$, and so every subset of $X \times [0,1]$ of diameter $< \lambda$ is contained in some element of $\mathcal U$.
Let $N$ be a natural number with $N > \frac{1}{\lambda}$.
For each $x \in X$ and each $i=1,\ldots,n$, note that $\{x\} \times [i-1/n,i/n]$ has diameter $<\lambda$ and so $H\left(\{x\} \times [(i-1)/n,i/n]\right)$ is contained in some element of $\cal V$, and hence $f(x,(i-1)/n)$ and $f(x,i/n)$ have distance $\le 2 \cdot (\epsilon/2) = \epsilon$.
Now take $f_i(x) = f(x,i/n)$ for $i=0,\ldots,N$.
