Is the interval [0, 1] in homotopy subset of real number? I am asking this question within the framework of my earlier posting here, for which I am grateful to @aes and @kobe for their generous help, thank you! Here are the question and its solution again for your convenience:

QUESTION: 
Given that $h_0, h_1: X \to Y$ and $k_0, k_1: Y \to Z$ are homotopic, prove that $k_0 \circ h_0$ and $k_1 \circ h_1$ are homotopic.

SOLUTION: 
(i) Here we are given that 
$$k_0 \circ h_0: X \to Z,\\
k_1 \circ h_1: X \to Z.$$
(ii) To prove that $k_0 \circ h_0 \simeq k_1 \circ h_1$, I claim that there exists a map $L: X \times I \to Z$, such that $L(x, 0) = k_0 \circ h_0 (x)$ as well as $L(x, 1) = k_1 \circ h_1(x)$, where $I = [0, 1]$ and $x \in X.$
(iii) And according to the response I got, that map $L$ would either be:
$$\begin{align}
L(x, t) &= k_t (h_t(x)), \quad t \in [0, 1], \text{ or }\tag{iiia}\\
\\
L(x, t) &= \tag{iiib}
\begin{cases}
k_0 (h_{2t} (x) ), &&\text{ for } t \in[0, \frac12]\\
k_{2t-1} (h_{1} (x) ), &&\text{ for } t \in [\frac12, 1].
\end{cases}
\end{align}$$

Here are my burning questions: 
(1) Is the interval $I = [0, 1]$ subset of real number $\mathbb R, \mathbb Q$ or simply has to be read as either $0$ or $1$ only?
(2) If $I \subset \mathbb R$ then (iiia) and (iiib) will be in trouble since, for example, for $t = 0.3$ the $L (x, 0.3)$ will be a joker. I am pretty sure this is a fool's question but I don't know where the dummy is hiding.
(3) Do you think this boring, unimaginative map may count as an answer too, if $I \subset \mathbb R$? 
$$L(x, t) =
\begin{cases}
k_0 \circ h_0(x), &t \in [0, \frac12)\\
k_1 \circ h_1(x), &t \in [\frac12, 1]
\end{cases}$$
Note the half-open interval $[0, \frac12)$ in the first piece.
(4) And finally, how about this plain vanilla $L(x, t)$, regardless of $I$ is a subset $\mathbb R$ or not:
$$L(x, t) = (1-t)k_0 \circ h_0(x) + (t)k_1 \circ h_1 (x)?$$
Thank you very much for your time and help.

POST SCRIPT: I pasted this image as an exhibit to my response to Carlos Laguillo's answer, see his answer below. The complete page can be found in this link here. The question I posted above is Munkres' $\S$ 51 Exercise #1, the one I pasted below is $\S$ 51 Exercise #2. Many thanks to Carlos.
$$\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}$$

 A: I would have left this as a comment, but I haven't enough reputation (I'm new here).
I think this is a problem on notation. In order to make this has sense and be correct we must understand $k_{t}$ as $H_{k}(x,t)$ where $H_{k}(x,t)$ is the homotopy between $k_{0}$ and $k_{1}$, and similarly for $h_{t}$. Then both answers (iiia) and (iiib) are correct.
Answering your questions:
(1) $I\subset \Bbb{R}$, we can think on it as the parameter that determines the continuous deformation from one map to another.
(2)This is the notation problem I mentioned above.
(3)No, that map doesn't need to be continuous on $t$. In fact, it will be continuous iff $k_{0}\circ h_{0}=k_{1}\circ h_{1}$.
(4)You were answered about this one in the other post, you can't multiply in general topological spaces, it may not make any sense.
A: Let me start off with a formal proof of the statement in question, then I'll get to your questions.
Claim: Let $h, h' : X \to Y$ and $k, k' : Y \to Z$. If $h \simeq h'$ and $k \simeq k'$, then $k \circ h \simeq k' \circ h'$.
Proof: Since $h \simeq h'$, there exists a continuous $H : X \times I \to Y$ with $H(s, 0) = h(s)$ and $H(s, 1) = h'(s)$. Define $h_t : X \to Y$ by $h_t(s) = H(s, t)$ and notice that
\begin{align*}
h_0(t) &= H(s, 0) = h(s) \\
h_1(s) &= H(s, 1) = h'(s).
\end{align*}
Since $k \simeq k'$, there exists a continuous $K : Y \times I \to X$ with $K(s, 0) = k(s)$ and $K(s, 1) = k'(s)$. Define $k_t : Y \to Z$ by $k_t(s) = K(s, t)$ and notice that
\begin{align*}
k_0(s) &= K(s, 0) = k(s) \\
k_1(s) &= K(s, 1) = k'(s)
\end{align*}
Define $L : X \times I \to Z$ by $$L(s, t) = \begin{cases} k_0\big(h_{2t}(s) \big) & 0 \leq t \leq 1/2 \\ k_{2t - 1} \big( h_1(s) \big) & 1/2 \leq t \leq 1.\end{cases}$$ Observe that
\begin{align*}
L(s, 0) &= k_0 \big( h_0(s) \big) = k \circ h(s) \\
L(s, 1) &= k_1 \big( h_1(s) \big) = k' \circ h' (s).
\end{align*}
Now notice that $ k_0\big(h_{2t}(s) \big)$ and $ k_{2t - 1} \big( h_1(s) \big)$ are continuous on $[0, 1/2]$ and $[1/2, 1]$, respectively, since the composition of continuous functions is continuous. Moreover they both agree at $t = 1/2$. By the pasting lemma, we can conclude that $L$ is continuous and hence the desired homotopy.

Let's start with looking at your two possible definitions of $L$. The problem with (iiia) is: how do you show it's continuous? Obviously $L(s, 0) = k \circ h(s)$ and $L(s, 1) = k' \circ h'(s)$, but how do you get continuity? You have two functions in two variables both changing at the same time which may cause you to work more to show continuity.
This is why you use (iiib). The continuity follows from the fact that we are composing a function in one variable with a function in two variables, for instance $k_0 \circ h_{2t}(s)$; the $k_0$ is fixed, unlike $k_t \circ h_t$ where the $k_t$ is changing as $t$ changes. We break it up into pieces that are continuous (pieces we can easily show to be continuous) and then use the pasting lemma to show the entire function you've just defined is continuous.
Edit: Disregard these last two paragraphs. Both functions are continuous by the same reasoning: the composition of two continuous functions is continuous. As long as we've shown $k_t$ and $h_t$ are continuous in two variables, both your definitions of $L$ should work perfectly. See here.
For (1), $[0, 1]$ is the closed set of all elements in $\mathbb R$ between $0$ and $1$, hence there are uncountably infinitely many.
For (2), you don't have any problems with $t = 0.3$. since $L$ is defined on $X \times I$ and $0.3 \in I$.
For (3), this is not well-defined as $t$ doesn't play a role at all and $L$ must be defined on $X \times I$, not just $X \times \{0, 1\}$ (by definition of homotopy).
For (4), the definition doesn't work because scalar multiplication is not necessarily defined, but also, you're not given that the space is convex, so even if scalar multiplication were defined, the function may not be well defined if it were not convex.
