What does "definition is independent of the choice of" mean?
An example:
Let $W$ be Banach and $V$ a normed space. Let $X$ be a dense subspace of $V$. Let $T \in Lin(X,W)$. For every $v \in V$ there exists $(x_k) \in X$ s.t. $\lim_{k \rightarrow \infty} x_k = v$.
Prove that the definition of $\lim_{k \rightarrow \infty} T(x_k)$ is independent of the choice of the sequence $(x_k)$.
On
To gain intuition, start with a simpler example: in basic algebra the slope of a line is often defined by a calculation based on 2 distinct but otherwise arbitrary points $(x_1, y_1), \, (x_2,y_2)$ which lie on the line:
$$ m = \frac{y_2 - y_1}{x_2 - x_1}.$$
In order for you to be able to describe the resulting quantity as being the slope of the line rather than merely a slope of the line, you need to show that this definition of slope is independent of the choice of the two points. This would mean that if $(x'_1, y'_1), \, (x'_2,y'_2)$ are another pair of distinct points on the line then
$$ \frac{y'_2 - y'_1}{x'_2 - x'_1} = \frac{y_2 - y_1}{x_2 - x_1}$$
(as verified by an easy argument with similar triangles).
On
Another example: modular arithmetic. The sum of congruence classes mod $m$ can be defined as $$\overline{a} + \overline{b} = \overline{x + y}$$ for any $x\in\overline{a},$ $y\in\overline{b}$, because $\overline{x + y}$ is independent of the choice of the representatives $x,y$.
It means that if you take another object satisfying assumptions then what comes out is the same result. Therefore we can define something without specifying the underlying object (any is good) which might be technically necassary for the definition.
In your example: if you take another sequence $(y_k)\subseteq X$ such that $\lim y_k=v$ then $$\lim T(y_k)=\lim T(x_k)$$
(The equality is what you have to prove)
Therefore you can define $T(v):=\lim T(x_k)$. This definition is correct because of the independence (the continuity and/or linearity of $T$ is a different problem). I don't have to specify how exactly I constructed the sequence $(x_k)$. Any is good.