My conjecture is that a continuous function is invertible only if it is strictly monotonic.
I assume that this is true, because if a continuous function $f$ is not strictly monotonic, then there must be a point $b$ where for some point $a<b$, and for some point $c>b$, we have $f(a)<f(b)$ and $f(c)<f(b)$. Then because of the intermediate value theorem, there must be some point $d$, where $a<d<b$, and some point $e$, where $b<e<c$, such that $f(d)=f(e)$. (or the inequalities are flipped) In other words, if it is not strictly monontonic, then $f$ must go down after having gone up (or the other way around), and therefore there must be outputs that share the same input.
Is this correct?
For precisely the reasons you state, an arbitrary function $f:\mathbf{R}\to \mathbf{R}$ which is strictly monotonic will be injective. Indeed, if $f(x)=f(y)$, then we must have that $x=y$, else $x>y$ and $f(x)>f(y)$ or $x<y$ and $f(x)<f(y)$. Notice how this doesn't need continuity. This does give us left invertibility, i.e. there exists a function $g$ so that $g\circ f=\text{Id}$. We might ask, however, when we can get that our function is invertible in the stronger sense - i.e., when our function is a bijection.
If we promote our function to being continuous, by the Intermediate Value Theorem, we have surjectivity in some cases but not always. Suppose that $f$ is strictly increasing, since the case of strictly decreasing is basically the same argument. Suppose that for all $M\in \mathbf{R}_{>0},$ we can find an $x$ sufficiently large so that $f(x)>M$. Suppose, in addition, that we can find a $z$ sufficiently small (i.e. negative) so that $f(z)<-M$. Then, we know that $[-M,M]\subset f(\mathbf{R})$ for all $M\in \mathbf{R}_{>0}$, by the IVT. This implies that $f$ is surjective, and injective, and hence a bijection.
Notice how $f:\mathbf{R}\to \mathbf{R}$ can be continuous and strictly increasing without being surjective. Take, for example, $f:\mathbf{R}\to \mathbf{R}$ given by $f(x)=\arctan(x)$.