I'm trying to understand a proof of a lite version of the van Kampen theorem. It says
Let $x\in X$ and $U,V\subset X$ open subsets containing $x$ s.t. $U\cap V$ is path connected. Let $U\rightarrow X$, $V\rightarrow X$ be inclusion maps. Then $\pi_1(X,x)$ is generated by the images of $\pi_1(U,x)\rightarrow \pi_1(X,x)$ and $\pi_1(V,x)\rightarrow \pi_1(X,x)$.
In the proof $\alpha\in \pi_1(X,x)$ we want to find a presentation $\alpha=[\lambda]$ for the concatenation $\lambda=\lambda_1*\ldots*\lambda_{2m}$ where $\lambda_j$ are loops in $(X,x)$ and $\lambda_{2i}(I)\subset V$ and $\lambda_{2i-1}(I)\subset U$.
The first part of the proof states:
To find a concatenation we shall use compactness to find positive integer $n$ s.t. for each $i$ the subinterval $[\frac{(i-1)}{n},\frac{i}{n}]$ is contained in either $\lambda^{-1}(U)$ or $\lambda^{-1}(V)$. Let $\lambda_i'(t)=\lambda(\frac{i-1+t}{n})$ so that $\lambda=\lambda_1'*\ldots*\lambda'_{n}$. After possibly concatenating with constant paths on the left and on the right we may assume $\lambda_1'(I)\subset U$ and $\lambda_n'(I)\subset V$.
I'm a little confused about this. Why do we conclude that $[\frac{(i-1)}{n},\frac{i}{n}]$ is contained in either $\lambda^{-1}(U)$ or $\lambda^{-1}(V)$? I don't really understand the part with "After possibly concatenating..."
Going further in the proof we show that any concatenation $\lambda_1'*\ldots*\lambda_n'$ with $\lambda_1'(I)\subset U$, $\lambda_n'\subset V$ and the rest in either $U$ or $V$, is homotopic relative to concatenation $\lambda=\lambda_1*\ldots*\lambda_{2m}$ described above. This seems fine to me.
Now for the last part of the proof, we need to show that $\lambda_i$ indeed are loops. The proof states it follows from the fact that $x_i=\lambda_i(1)=\lambda_{i+1}(0)\in U\cap V$ is path connected. This makes sense because we then can find paths from $x$ to $x_i$ and replace the original with loops. However how do we know that there is an $x_i\in U\cap V$ s.t. $x_i=\lambda_i(1)=\lambda_{i+1}(0)$?
As Ted Shifrin pointed out, the concept of a Lebesgue number is the key to the proof presented in your question. Given a metric space $(X,d)$ and an open cover $\mathcal{W}$ of $X$, a Lebesgue number for $\mathcal{U}$ is a real number $\rho > 0$ such that each $M \subset X$ with diameter $< \rho$ is contained in some $W \in \mathcal{W}$. It is a well-known theorem that each open cover $\mathcal{W}$ of a compact metric space $(X,d)$ has a Lebesgue number. For a proof see any book or my answer to Relation between Riemann sums and oscillation of a bounded function (note that it suffices to consider open covers of the special form $\{ B(x,\delta(x)) \}_{x \in X}$ with $\delta(x) > 0$ and $ B(x,\rho) = \{ y \in X \mid d(x,y) < \rho \}$).
Now $I$ with the metric $d(x,y) = \lvert x - y \rvert$ is a compact metric space and $\mathcal{W} = \{ \lambda^{-1}(U), \lambda^{-1}(V) \}$ is an open cover of $I$. Let $\rho$ be a Lebesgue number for $\mathcal{W}$ and $n \ge 1/\rho$. Then for each $i$ the subinterval $[\frac{(i-1)}{n},\frac{i}{n}]$ is contained in $\lambda^{-1}(U)$ or $\lambda^{-1}(V)$. This gives you the paths $\lambda'_i$ such that $\lambda = \lambda'_1 * \dots * \lambda'_n$. Unfortunately the paths $\lambda'_i$ are not necessarily closed. But $U \cap V$ is path connected, thus for each $i = 1,\dots,n-1$ we can find a path $\mu_i$ in $U \cap V$ from $\lambda'_i(1) = \lambda'_{i+1}(0)$ to $x$. Define $\lambda''_1 = \lambda'_1 * \mu_1$, $\lambda''_n = \mu_{n-1}^{-1} *\lambda'_n$ and $\lambda''_i = \mu_{i-1}^{-1} *\lambda'_i * \mu_i$ for $i = 2,\dots,n-1$. Then each $\lambda''_i$ represents an element of in the image of $\pi_1(U,x)\rightarrow \pi_1(X,x)$ or $\pi_1(V,x)\rightarrow \pi_1(X,x)$ and we have $$\lambda = \lambda'_1 * \dots * \lambda'_n \simeq \lambda'_1 * \mu_1 * \mu_1^{-1} *\lambda'_2 * \mu_2 * \dots * \mu_{n-2}^{-1} * \lambda'_{n-1} * \mu_{n-1} * \mu_{n-1}^{-1} * \lambda'_n$$ $$= \lambda''_1 * \dots * \lambda''_n$$ where $\simeq$ denotes homotopy rel. $\{ 0, 1\}$.
This proves your claim. Further arguments are not needed. In particular, it is irrelevant whether $\lambda_1'(I)\subset U$ and $\lambda_n'(I)\subset V$ (but you may assume this if you want).