Edit: This answer is wrong. But it was a good try, if I may say so myself. I am leaving it up; maybe the idea is still usable somehow. (Put more wiggles in the functions? I don't know.)

No. Let $A$, resp. $B$, be the graph $x=f(y)$, resp. $x=g(z)$, for some continuous functions $f$ and $g$. Pick numbers $0<d<b<a<c<1$, and arrange things so that $f(0)=g(0)=0$, $f(1)=g(1)=1$, and $f(y)$ grows from 0 to $a$, then decreases to $b$, then finally increases to 1, whereas $g(z)$ grows from $0$ to $c$, decreases to $d$, and increases to 1. (Here, $a$, $b$, $c$, $d$ are function values, i.e., values of $x$, not of the arguments $y$ or $z$.)

Now if $C$ projects to $A$ and $B$, respectively, it is not hard to see that $(0,0,0)\in C$ and $(1,1,1)\in C$. Trying to move $(x,y,z)$ continuously from the former to the latter, a contemplation of how $x$ must grow and shrink in order to maintain $x=f(y)$ and $x=g(z)$ leads to a contradiction. (Hard to put in words, but a pair of graphs reveals it I think.)

Edit: This answer is wrong. But, since Anton based his answer on this idea, I am leaving it up.

No. Let $A$, resp. $B$, be the graph $x=f(y)$, resp. $x=g(z)$, for some continuous functions $f$ and $g$. Pick numbers $0<d<b<a<c<1$, and arrange things so that $f(0)=g(0)=0$, $f(1)=g(1)=1$, and $f(y)$ grows from 0 to $a$, then decreases to $b$, then finally increases to 1, whereas $g(z)$ grows from $0$ to $c$, decreases to $d$, and increases to 1. (Here, $a$, $b$, $c$, $d$ are function values, i.e., values of $x$, not of the arguments $y$ or $z$.)

Now if $C$ projects to $A$ and $B$, respectively, it is not hard to see that $(0,0,0)\in C$ and $(1,1,1)\in C$. Trying to move $(x,y,z)$ continuously from the former to the latter, a contemplation of how $x$ must grow and shrink in order to maintain $x=f(y)$ and $x=g(z)$ leads to a contradiction. (Hard to put in words, but a pair of graphs reveals it I think.)

No. Let $A$, resp. $B$, be the graph $x=f(y)$, resp. $x=g(z)$, for some continuous functions $f$ and $g$. Pick numbers $0<d<b<a<c<1$, and arrange things so that $f(0)=g(0)=0$, $f(1)=g(1)=1$, and $f(y)$ grows from 0 to $a$, then decreases to $b$, then finally increases to 1, whereas $g(z)$ grows from $0$ to $c$, decreases to $d$, and increases to 1. (Here, $a$, $b$, $c$, $d$ are function values, i.e., values of $x$, not of the arguments $y$ or $z$.)

Now if $C$ projects to $A$ and $B$, respectively, it is not hard to see that $(0,0,0)\in C$ and $(1,1,1)\in C$. Trying to move $(x,y,z)$ continuously from the former to the latter, a contemplation of how $x$ must grow and shrink in order to maintain $x=f(y)$ and $x=g(z)$ leads to a contradiction. (Hard to put in words, but a pair of graphs reveals it I think.)

Edit: This answer is wrong. But it was a good try, if I may say so myself. I am leaving it up; maybe the idea is still usable somehow. (Put more wiggles in the functions? I don't know.)

No. Let $A$, resp. $B$, be the graph $x=f(y)$, resp. $x=g(z)$, for some continuous functions $f$ and $g$. Pick numbers $0<d<b<a<c<1$, and arrange things so that $f(0)=g(0)=0$, $f(1)=g(1)=1$, and $f(y)$ grows from 0 to $a$, then decreases to $b$, then finally increases to 1, whereas $g(z)$ grows from $0$ to $c$, decreases to $d$, and increases to 1. (Here, $a$, $b$, $c$, $d$ are function values, i.e., values of $x$, not of the arguments $y$ or $z$.)

Now if $C$ projects to $A$ and $B$, respectively, it is not hard to see that $(0,0,0)\in C$ and $(1,1,1)\in C$. Trying to move $(x,y,z)$ continuously from the former to the latter, a contemplation of how $x$ must grow and shrink in order to maintain $x=f(y)$ and $x=g(z)$ leads to a contradiction. (Hard to put in words, but a pair of graphs reveals it I think.)