Let $$S = (0, 1) \cup (1, 2) \cup (3, 4)$$ and $$T = \{0, 1, 2, 3\}$$. Then which of the following statements is(are) true?

Let $$S = (0,1)\cup(1,2)\cup(3,4)$$ and $$T=\{0,1,2,3\}$$.

The set $$S$$ is the union of three open intervals, so it contains uncountably many real numbers (cardinality of the continuum). The set $$T$$ is finite with $$|T|=4$$.

Option A - “There are infinitely many functions from $$S$$ to $$T$$”.
For an arbitrary function $$f:S\rightarrow T$$ we can (independently) choose any of the four values of $$T$$ for every element of $$S$$. Hence the total number of such functions is $$4^{|S|}$$. Since $$|S|$$ is uncountable, $$4^{|S|}$$ is also uncountable, in particular infinite. Therefore Option A is true.

Option B - “There are infinitely many strictly increasing functions from $$S$$ to $$T$$”.
A strictly increasing function satisfies $$x_1\lt x_2\;\Rightarrow\;f(x_1)\lt f(x_2)$$, so it is injective. Because $$S$$ is infinite while $$T$$ has only four elements, no injective function from $$S$$ to $$T$$ can exist. Thus there is not even one (let alone infinitely many) strictly increasing function. Option B is false.

Option C - “The number of continuous functions from $$S$$ to $$T$$ is at most 120”.
The three components $$(0,1),\,(1,2),\,(3,4)$$ of $$S$$ are connected sets (intervals). The image of a connected set under a continuous function is also connected. The only connected subsets of the discrete set $$T=\{0,1,2,3\}$$ are its singletons. Hence a continuous function $$f:S\rightarrow T$$ must be constant on each of the three intervals:

$$f(x)=a\;\text{on }(0,1),\qquad f(x)=b\;\text{on }(1,2),\qquad f(x)=c\;\text{on }(3,4),$$

where $$a,b,c\in T$$. Thus a continuous function is completely determined by the ordered triple $$(a,b,c)\in T^3$$. The total number of such triples is $$|T|^3 = 4^3 = 64\le 120$$. Therefore Option C is true.

Option D - “Every continuous function from $$S$$ to $$T$$ is differentiable”.
As shown above, every continuous $$f$$ is constant on each open interval. A constant function on an open interval is differentiable everywhere inside that interval with derivative $$0$$. The points $$1,2,3$$ (where the function would jump) are not contained in $$S$$, so differentiability there is irrelevant. Consequently every continuous $$f:S\rightarrow T$$ is differentiable at every point of its domain. Option D is true.

Hence the correct statements are:
Option A, Option C, Option D.