Show that a linear projection is continuous if it is continuous in 0 Here is a perfectly fine answer to the question:
Linear functional $f$ is continuous at $x_0=0$ if and only if $f$ is continuous $\forall x\in X$?
However, I am in a set and topology course and my professor proved this using sequences.
I am going over the proof and am unsure of one step.
Let X with $\lVert{X}\rVert$ and Y with $\lVert{Y}\rVert$ be normed vector spaces(Over ${\rm I\!R}$). Let T: X -> Y be a linear projection.
Show that if T is continuous in $\overline{0}$ it is continuous*.
Solution:
Let a $\in$ X. Let $x_{n}$ -> a.
$x_{n}$ - a -> $\overline{0}$ $\quad$ Because X is a vector space and $x_{n}$ converges to a.
T($x_{n}$ - a) -> $\overline{0}$ $\quad$ **
T($x_{n}$) - T(a) -> $\overline{0}$ $\quad$ b/c T is a linear projection
T($x_{n}$) -> T(a) $\quad$ Which implies T is continuous in a.
*In every point I suppose
** Why is this ok? 
Is there some clever rule I'm missing?
/Regards 
 A: What your techer used were sequences, not series.
The proof is correct. And it doesn't apply to $T(x)=x+b$ since this is not a linear map.
A: First off, recall that a function $f:X\rightarrow Y$ over a metric spaces $(X,d), (Y,d')$ is continuous at $a \in X$ if and only if given any sequence $(x_n)_{n\in\mathbb{N}} \subseteq X$ convergent to $a$, it holds that
$$
f(a) = f(\lim_{n \to \infty} x_n) = \lim_{n \to \infty}f(x_n)
$$
Now, to answer $(*)$, the correct statement is

Given $T$ a linear operator in a normed space $(E, ||\cdot||)$, T is continuous (that is, at every point, or equivalently, that it holds the previously said) if and only if it is continuous at $0$.

Regarding $(**)$, this is because since $(x_n)_{n \in \mathbb{N}}$ converges to $a$, $(x_n - a)_{n \in \mathbb{N}}$ converges to zero because, trivially,
$$
||x_n - a|| = ||(x_n - a) - 0||
$$
Now, by continuity and using the characterization that I've previously stated, $T(x_n -a)$ converges to zero because $T$ is continuous at zero, and by linearity, this is the same as $T(x_n) - T(a)$ converging to zero, so 
$$
T(a) = \lim_{n \to \infty} T(x_n)
$$
and again by the characterization of continuity that we have, this implies that $T$ is continuous at $a$.
