Epsilon-Delta in the limit of a function

Question

In class we are busy working with the limit of a function. I have been shown a necessary and sufficient sketch proof, yet I don't understand the change of inequality for the limit of the sequence. I am struggling and more worried about understanding the $\implies$ direction.

1. Suppose $f$ is continuous at $x$ and let $\{x_k\}_{k=1}^\infty$ be any sequence converging to $x$.
2. Fix $\epsilon>0$. By continuity of $f$ at $x$, there exists $\delta$ such that
3. $|x^{'}-x|<\delta \implies |f(x^{'})-f(x)|<\epsilon$ .
4. Since $x_k\rightarrow x $, there exits $K$ such that for all $k>K$ we have $|x_k-x|<\delta$.
5. We conclude that for all $k\geq K$, we have $|f(x_k)-f(x)|<\epsilon$

So, I am struggling with understanding line 4. As previously we have been given the definition of the limit of a sequence being,

$ \exists K \in \mathbb{N} s.t. \forall k>K, \ |x_k-x|<\epsilon $

Did we get from $|x_k-x|<\epsilon$ to $|x_k-x|<\delta$ because $\delta<\epsilon $?

P.S. This is one of my first posts on stackexchange, so any feedback would be great.