Intermediate Value Theorem in $\mathbb{R}^2$ and $\mathbb{R}^3$

Let $$f:\mathbb{R}^n \rightarrow \mathbb{R}$$ be a continuous function and let $$a,b \in \mathbb{R}^n$$. Let $$g:\mathbb{R} \rightarrow \mathbb{R}$$ be defined as:

$$g(t) = f(ta + (1-t)b)$$.

i) Prove that $$g$$ is continuous.

My answer: I have done this using the fact that g is a composition of continuous functions and the fact that f is continuous

ii) Prove that the following Intermediate Value Theorem is true for $$f$$:

If $$d$$ is a number between $$f(a)$$ and $$f(b)$$ then there exists a point $$c \in \mathbb{R}^n$$ such that $$f(c)=d$$.

My attempt at a proof:

First we make the observation that $$g(0)=f(b)$$ and that $$g(1)=f(a)$$.

Since $$g$$ is a real and continuous function $$\implies$$ there exists a $$c' \in[0,1]$$ such that $$g(c')=d$$ since d is a number between $$f(a)$$ and $$f(b)$$. Thus $$f(c'a+(1-c')b)=d$$ and were are done.

iii) State and prove a similar Intermediate Value Theorem for continuous functions $$h:R \rightarrow \mathbb{R}$$ and $$h':B \rightarrow \mathbb{R}$$ where $$R$$ is an open or closes rectangle in $$\mathbb{R}^2$$ and B is an open or closed ball in $$\mathbb{R}^3$$

My thoughts: Since we have proven it for $$\mathbb{R}^n$$ is the arguement not just to say that both sets are subsets of $$\mathbb{R}^n$$?

I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!

No, you have to do a little bit more, because for given $$a,b\in R$$ and $$d$$ between $$h(a)$$ and $$h(b)$$, you have to check that $$c$$ is still in $$R$$ and not in $$\mathbb R^2\setminus R$$. But using the same ansatz as above, you have just to do a slight change to get a $$c\in R$$ such that $$h(c)=d$$.
Your proof i) and ii) is correct, but at ii) you should define $$c:=c'a+(1-c')b$$ at the end to get the $$c$$.