Doubt on a proof of continuity of a monotonically increasing function on an interval

Question

Consider the following statement:
Let $f:I \to R$ be an increasing function. Let $c\in I$ not be an endpoint of the interval $I$. Then prove that $f$ is continuous at $c \in I\; $ if and only if $\; \exists $ a sequence $(x_n) \subset I$ such that $x_{2n-1} \lt c \;\; \text{and} \; x_{2n} \gt c \;\forall n\in N; \text{and such that} \;x_n \to c \; \text{and}\; f(x_n)\to f(c)$

I have tried to prove the above statement as follows:

To prove: $f$ is continuous at $c \in I\; \text{(not an endpoint)} \iff \exists $ a sequence $(x_n) \subset I$ such that $x_{2n-1} \lt c \;\; \text{and} \; x_{2n} \gt c \;\forall n\in N; \text{and such that} \;x_n \to c \; \text{and}\; f(x_n)\to f(c)$ $(P \iff Q, \text{say})$
$P\Rightarrow Q$: trivial
$P \Leftarrow Q:$
If for $x\in I, x\lt c\;\text{then,}\\ f(x)\le f(c) \Rightarrow sup f(x)\le f(c)$
Since $f$ is increasing hence $\lim_{x\to c^-} f(x)= \sup(f(x): x<c)=\lim (x_{2n-1})=f(c) \; \text{(using sequential criteria for $\lim_{x\to c^-} f(x))$ and that $f(c)=\lim f(x_n))$} $
Similarly, $\lim_{x\to c^+} f(x)=\inf(f(x): x>c)=\lim f(x_{2n})=f(c)$
Since, $\lim_{x\to c^-} f(x)=\lim_{x\to c^+} f(x)=f(c) \Rightarrow f \text { is continuous at c}$. Hence Proved.

Is my proof correct?
I think it is wrong because it may so happen that the sequence $(x_n)$ defined in $(Q)$ does not exist however if for all the sequences $(y_n)$ which exist and converge to $c, f(y_n)$ also converges to $f(c)$, then by sequential criteria for continuity of a function, $f$ is continuous at $c$.
My question is how do I prove that the existence of $(x_n)$ as defined in $Q$ above is sufficient for $f$ to be continuous at c? Please advice what I am missing here. Thanks in advance.