It is a funny countable vs uncountable argument, I least expected that. My hint (in the comments) to use a nested sequence of intervals only made me confused. (But it made me think that if $C$ is the union of all open intervals on which $f$ is locally constant then $f(C)$ is countable. I tinkered a bit with ideas like considering cases, whether $[0,1]\setminus C$ contains an open interval (then the proof by the OP works), or whether $C$ is dense, looked at $f([0,1])\setminus f(C)$ and at some point figured it was easier than that and I may formally forget about $C$, and ignore some irrelevant details.)
So say $f$ were not constant, hence $f([0,1])=[m,M]$ with $m<M$. For each $y\in[m,M]$ pick $x_y\in[0,1]$ with $f(x_y)=y$. Let $X=\{x_y:y\in[m,M]\}$. In other words we picked a set $X\subset[0,1]$ such that the restriction of $f$ to $X$ is one-to-one and onto $[m,M]$.
Let $T=\{x\in X: f $ has a local maximum at $x\}$ and $S=\{x\in X: f $ has a local minimum at $x\}$. Since $X=T\cup S$ and (cardinality) $|X|=|[m,M]|=\mathfrak c = 2^{\aleph_0}$, at least one of $T$ and $S$ is uncountable. Say $T$ is uncountable.
For each $x\in T$ pick $n_x\in\mathbb N$ such that $f(x)\ge f(z)$ whenever $z\in(x-\dfrac1n,x+\dfrac1n)\cap[0,1]$. Let $T_n=\{x\in X: n_x=n\}$. Clearly $T=\bigcup\limits_{n\in\mathbb N} T_n$, hence there is an $n$ for which $T_n$ is uncountable.
Then $T_n$ must have a limit point, i.e. a $t\in[0,1]$ such that every neighborhood of $t$ contains infinitely many elements of $T_n$. Hence we may pick two different points $p,q\in T_n$ each very close to $t$ and hence (absolute value) $|p-q|<\dfrac1n$. But this means $f(p)\le f(q)$ and $f(q)\le f(p)$, thus $f(p)=f(q)$, contradicting that the restriction of $f$ to $X$ were one-to-one.
The function $\Delta : (x_1,x_2,y) \mapsto (\phi(x_1,y),\phi(x_2,y))$ is continuous iff the two functions $f:(x_1,x_2,y) \mapsto \phi(x_1,y)$ and $g:(x_1,x_2,y) \mapsto \phi(x_2,y)$ are continuous.
To see that this is the case, notice that $f$ is the composition of $\phi$ and $(x_1,x_2,y) \mapsto (x_1,y)$ and that both are continuous.
Best Answer
I think this might be a rather geometric proof.
Consider a continuous function $f\colon [a,b]\to \mathbb{R}$. Let $\ell$ be the line that goes through the points $(a,f(a))$ and $(b,f(b))$.
If the graph of $f$ is equal to $\ell$ (or a segment of it, to be more precise), then there is nothing to prove. Note that in this case $$f(x) = \frac{f(b)-f(a)}{b-a}(x-a) +f(a).$$
If the graph of $f$ is not equal to the line $\ell$, then there is one point $(x,f(x))$ on the graph of $f$ that does not lie on the line segment.
Consider a line $\tilde \ell$ that is parallel to the line $\ell$ and lies "between" the point $(x,f(x))$ and the line $\ell$ (i.e. $\tilde \ell$ seperates the point $(x,f(x))$ and the line $\ell$). Now, since $f$ is continuous and the interval $[a,b]$ is connected, the graph of $f$ is also connected. Therefore, the graph of $f$ must intersect $\tilde \ell$ at a point $(x_1,f(x_1))$ with $x_1 \in (a,x)$ and also at another point $(x_2,f(x_2))$ with $x_2 \in (x,b)$. Since these two points lie on $\tilde \ell$, we have $$ \frac{f(b)-f(a)}{b-a} = \frac{f(x_2)-f(x_1)}{x_2-x_1}.$$