A function $f$ has period $t$ if for all $x$ in the domain it is true that $f(x+t) = f(x)$. A function is called periodic if it has (at least one) period. Take any periodic function $f$ -periodicity does not necessarily implies continuity- and define its fundamental period $T=T(f)$ as $T=\inf \{t>0:$ t is a period of $f\}$. My purpose is to show that if $T>0$ then it must be the case that $T$ is a period of $f$.
The approach I took is to take a strictly decreasing sequence $(\epsilon_n)_{n\in\mathbb{N}}$ such that $\epsilon_n > 0 \ \forall \ n$ and $\epsilon_n \rightarrow 0$. By definition of infimum we must have a sequence of periods for $f$, $(t_n)_{n\in\mathbb{N}}$ that converges to $T$ from which you can assume without loss of generality that $t_n \geq t_{n+1} \geq T$ for all $n\in\mathbb{N}$. Taking a fixed $x$ in the domain of the function we have that: $$|f(x+T)-f(x)| \leq |f(x+T)-f(x+t_n)| + |f(x+t_n) - f(x)| = |f(x+T)-f(x+t_n)|.$$ If we also had continuity then this would suffice. Or even continuity from the right, right? I know periodic functions do not allow all types of discontinuities, but I am not quite familiar on which specific type of discontinuities they allow. Could anyone care to explain? I am not quite sure my argument is sufficient for the proof.
Your proof is right in the continuous case only, but I think it can be done far more elegantly; in particular, let $T$ be the set of periods $t$ such that $f(x)=f(x+t)$ for all $x$. Notice that if $x\in T$, then so is $-x\in T$. Further, if $x,y\in T$, then $x+y\in T$. Finally, $0\in T$. These are easily verified. This means that $T$ is a group under addition. Thus, we can either say that $T$ is dense, or is the set $\{nP:n\in N\}$ where $P$ is the fundamental period - to prove this, suppose that $P>0$ were the infimum of the positive elements of $T$, but that there was some $k\in T$ which was not a multiple of $P$. Let $n=\lceil\frac{k}P\rceil$. We can easily show that, if $P$ is the infimum of the positive elements of $T$, then there is element $y$ of $T$ in the interval $[P,P+\frac{P-nP+k}n)$, since $\frac{P-nP+k}n>0$, implying $ny\in[nP,k+P)$ and that $ny-k\in(0,P)$, but since $ny-k$ is in $T$, this would imply there is a positive element less than $P$, thus $P$ couldn't be the infimum - a contradiction. Therefore $P$ must divide every $k\in T$ and from here, it follows that $T$ must be exactly the set of multiples of $P$.
Also, periodic is a very weak condition in terms of continuity; essentially, you get to choose any function $[0,P)\rightarrow \mathbb{R}$ and just tile it by periodicity - so things like the Dirichlet function (mapping rationals to 1 and irrationals to 0) are periodic, but discontinuous everywhere. Indeed, since an open interval is contained in $[0,P)$, wherein we choose our function without constraint, any type of discontinuity is allowable.
A more elegant proof of the property would be to, using the same terminology as before, notice that there must be no elements in $(0,P)$ and $T$. Thus, for any $x\in T$, it follows that $(x,P+x)$ is disjoint from $T$. But this implies that if $x\in (y,P+y)$, then $y$ cannot be in $T$, as that would imply $x\not\in T$. Thus, the set of such $y$, which is $(x-P,x)$ must also be disjoint from $T$.
Since $P$ is the infimum, there must be some $x\in T$ in the range $P\leq x < 2P$. Then, the interval $(x-P,x)$ would have to be disjoint from $T$. Since $(0,P)$ overlaps that interval and is also disjoint, their union, $(0,x)$ must be disjoint. Thus $x$ would be the minimum value in $T$ and therefore equal $P$, the infimum.
This easily implies that $nP\in T$ for any $n\in\mathbb{Z}$, so the intervals $(nP,(n+1)P)$ are empty, and hence $T$ is exactly the set of $nP$.