Cauchy condition for functions

Suppose that $(x_n)$ converges to $a$. By the Bolzano-Weierstrass theorem, we know that the sequence $(f(x_n))$ must have a monotone subsequence $(f(x_{n_j}))$.

Further, $(f(x_{n_j}))$ must be bounded: taking $\epsilon=1$, there exists $\delta>0$ so that $\lvert x-a\rvert<\delta$ and $\lvert y-a\rvert<\delta$ implies $\lvert f(x)-f(y)\rvert<1$; in particular, for all $x\in(a-\delta,a+\delta)$ we have $\lvert f(x)-f(a+\frac{\delta}{2})\rvert<1$, which implies $$ \lvert f(x)\rvert<\lvert f(a+\tfrac{\delta}{2})\rvert+1\text{ for all }x\in(a-\delta,a+\delta); $$ since $x_{n_j}$ is eventually contained in $(a-\delta,a+\delta)$, the sequence $(f(x_{n_j}))$ is then bounded.

So, $(f(x_{n_j}))$ is bounded and monotone, and therefore converges to some $L$. We claim that $(f(x_n))$ must converge to $L$ as well.

Let $\epsilon>0$ be given. By assumption, there exists $\delta>0$ such that $0<\lvert x-a\rvert<\delta$ and $0<\lvert y-a\rvert<\delta$ implies $\lvert f(x)-f(y)\rvert<\frac{\epsilon}{2}$. Because $x_n\rightarrow a$, there exists $N\in\mathbb{N}$ so that $n>N$ implies $\lvert x_n-a\rvert<\delta$. Because $x_{n_j}\rightarrow a$ and $f(x_{n_j})\rightarrow L$, there exists $J\in\mathbb{N}$ such that $\lvert x_{n_J}-a\rvert<\delta$ and $\lvert f(x_{n_J})-L\rvert<\frac{\epsilon}{2}$.

Then for $n>N$, $$ \lvert f(x_n)-L\rvert\leq\lvert f(x_n)-f(x_{n_J})\rvert+\lvert f(x_{n_J})-L\rvert<\epsilon. $$ So, $f(x_n)\rightarrow L$, as claimed.