Prove that $\forall m \in \mathbb{Z}: f(x+m)=f(x) \ \forall x \in \mathbb{R}$

Question

I tried solving this problem for quite some time now but I think there must be a simple solution I am just not seeing so I would appreciate some kind of hint how to proceed.
Problem: Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and $f(x+1) = f(x) \ \forall x \in \mathbb{R}$. Show that $\forall m \in \mathbb{Z}: f(x+m)=f(x) \ \forall x \in \mathbb{R}$.
What I've got so far: I tried proving this via induction over $m \in \mathbb{N}$.
For $m=0$ we have $f(x+0)=f(x)$ which is obviously true. For $m=1$ the expression $f(x+1) = f(x)$ is true by definition.
Induction hypothesis: Assume $f(x+m)=f(x)$ or $f(x+(-m)) = f(x)$ to be true for an arbitrary $m$ and $\forall x \in \mathbb{R}$.
Induction step: $m \to m+1$
$$ f(x+(m+1)) = f(\underbrace{(x+1)}_{x'}+m) = f(x'+m) \overset{IH}{=} f(x') = f(x+1) \overset{Def}{=} f(x). $$ Since we want to prove it for all whole numbers: $-m \to -(m+1)$
$$ f(x+(-(m+1))) = f(\underbrace{(x+(-1))}_{x'} + (-m)) = f(x'+(-m)) \overset{IH}{=} f(x') = f(x+(-1)) $$ And here is where I got stuck because I cannot simply apply the definition of the function here because it doesn't say anything about $f(x-1)$. I tried figuring out what it means though: We have $f(x+1)=f(x) \ \forall x \in \mathbb{R}$. So we can just define $x' = x-2$ and get $f(x'+1) = f(x') = f(x-2)$ but also $f(x-1) = f(x-2)$. But this seems to even contradict the statement I am trying to prove doesn't it?