1. Understand the sets

We have $S=(0,1)\cup(1,2)\cup(3,4), \qquad T=\{0,1,2,3\}.$

So the domain $S$ has infinitely many real numbers, while the codomain $T$ has only $4$ elements.

2. Option A: There are infinitely many functions from $S$ to $T$

A function from $S$ to $T$ assigns to each point of $S$ one of the $4$ values in $T$.

Since $S$ is an infinite set, the total number of functions from $S$ to $T$ is $4^{|S|},$ which is certainly infinite.

Hence, A is true.

3. Option B: There are infinitely many strictly increasing functions from $S$ to $T$

A strictly increasing function $f:S\to T$ must satisfy: if $x<y$ in $S$, then $f(x)<f(y).$

But $T=\{0,1,2,3\}$ has only $4$ distinct values.

Now note that $S$ contains infinitely many points. In particular, any interval such as $(0,1)$ contains infinitely many points. On a strictly increasing function, distinct inputs must give distinct outputs, because if $x<y$, then $f(x)<f(y)$, so $f(x)\neq f(y)$.

Thus a strictly increasing function on $S$ would need infinitely many distinct output values, but $T$ has only $4$ values.

So such a function cannot exist.

Hence the number of strictly increasing functions is $0$, certainly not infinitely many.

Therefore, B is false.

4. Option C: The number of continuous functions from $S$ to $T$ is at most 120

Let $f:S\to T$ be continuous.

Since $T=\{0,1,2,3\}$ is a discrete subset of $\mathbb{R}$, and each component of $S$ is an interval, $(0,1),\ (1,2),\ (3,4),$ any continuous function from a connected interval into a discrete set must be constant on that interval.

So on each of the three connected components, $f$ must be constant:

• one constant value on $(0,1)$,
• one constant value on $(1,2)$,
• one constant value on $(3,4)$.

Each constant can be chosen independently from $T$, which has $4$ choices.

Therefore total number of continuous functions is $4\times 4\times 4=4^3=64.$

Since $64\le 120,$ C is true.

5. Option D: Every continuous function from $S$ to $T$ is differentiable

From the previous step, every continuous function is constant on each interval component of $S$.

A constant function is differentiable everywhere on its interval, with derivative $0$.

Now $S$ does not include the points $1,2,3,4,0$, so we only need differentiability at points inside the open intervals $(0,1)$, $(1,2)$, and $(3,4)$. On each such interval, the function is constant, hence differentiable.

Therefore, every continuous function from $S$ to $T$ is differentiable.

So D is true.

6. Final conclusion

The true statements are: $\boxed{A,\ C,\ D}$

This matches the stored correct answer.