Why $\{ \lambda x + (1-\lambda){y}\;\lvert\; \lambda \in [0,1] \}$ represents the line segment between $x,y$?

Any point on the line passing through $x$ and $y$ is expressed as a linear combination of these points. The requirement on $\lambda$ insures that the points are in between $x$ and $y$, rather than outside the segment.

We use your (correct) definition of the line $\vec x(t)= \vec x_0 + t\vec{v}$, where $\vec v$ marks the direction of $l$. Note that $\textit{any}$ vector that has the same direction as $\vec v$ will do to give the set of points lying on $l$, because when you change the magnitude of $\vec v$ you are just reparameterizing $l.$

With the foregoing in mind, fix $\vec x_0,\vec y_0\in \mathbb R^n$ and take $\vec v=\vec y_0-\vec x_0.$ This makes sense because the vector difference $\vec y_0-\vec x_0$ is a vector in the same direction as $l.$

Therefore, a parametric equation of $l$ is $\vec x(t)=\vec x_0+t(\vec y_0-\vec x_0),$ which gives, on rearranging,

$\vec x(t)=t\vec y_0+(1-t)\vec x_0$. Now, as $t$ ranges from $0$ to $1$, $\vec x(t)$ goes from $\vec y_0$ to $\vec x_0$. So values ot $t$ between $0$ and $1$ correspond to points on the segment $\overline{\vec y_0 \vec x_0}.$

In fact, you can check that $\vec x_0$ and $\vec y_0$ split the line into three regions: $t<0;\ 0\le t\le 1;\ t>1$, which if you make a graph, are easily located on $l$.

Let consider

$$P(\lambda)=\lambda x + (1-\lambda){y}$$

then

$$P(0)=y, \quad P(1)=x$$

and

$$P(\lambda)-P(0)=\lambda (x-y)$$

therefore for $\lambda$ varing form $0$ to $1$, $P(\lambda)$ describes linearly all the ponts between $y$ and $x$.