Prove that taking the orthogonal projection of a vector is continuous Let $Y$ be a closed subspace of the Hilbert space $X$, and define $T:X\to Y$ as
$$Tx = Proj_Y x$$
Then I want to check that $T$ is continuous. So I did the following, I took a converging sequence $(x_n) \in X$ with limit $x$, then I want to prove that $(Tx_n) \to Tx$. To this end I considered that $Tx_i=y_i$ and $Tx=y$, therefore
$$||y_n-y||=||y_n-x_n+x_n-y|| <||y_n-x_n||+||x_n-y||<||y^{*}-x_n||+||x_n-y||$$
Now,since the last inequality holds for all $y^{*} \in Y$ we pick this point such that $||y^{*}-x_n||<\frac{\epsilon}{3}$, so we get
$$||y^{*}-x_n||+||x_n-y||<\frac{\epsilon}{3}+||x_n-x+x-y||<\frac{\epsilon}{3}+||x_n-x||+||x-y^{**}||$$
and by the same argument as above, but know with $x$ we finally get:
$$\frac{\epsilon}{3}+||x_n-x||+||x-y^{**}||<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$   
The thing is that I am not sure of my above proof, in the part of choosing those $y^{*}$ and $y^{**}$.
I was thinking to use the Pythagoras identity, but I don't know how.
Can someone help me to prove correctly the above result please?
Thanks a lot in advance. 

Note

Thm. Let $Y$ a closed subspace of the Hilbert space $X$. then, for  $x ∈ X$, there exists a unique $y_ 0 ∈ Y$ such that
$$||x − y_ 0 || ≤ ||x − y||$$
for all $y ∈ Y$ . This is, $y _0$ is the nearest vector in $Y$ to $x$. We call $y_0$ the orthogonal projection of $x$ in $Y$
 A: Hint (for a complete proof of the result).
I suppose that you define $Tx = Proj_Y x=p_Y(x)$ as the point at smallest distance of $x$ belonging to $Y$. By the Hilbert projection theorem, $p_Y(x)$ exists and is uniques as $Y$ is supposed to be closed.
Now you can prove that a point in $Y$ is equal to $p_Y(x)$ if and only if $$\mathcal{Re} \langle x-p_Y(x),y-p_Y(x) \rangle \le 0 \text{ for all } y \in Y.$$
Based on that, you can prove that for $u,v \in X$, you have $$\Vert p_Y(u)-p_Y(v)\Vert \le \Vert u - v \Vert \text{ for all } u,v \in X \tag{1}.$$ To do so, take $$z=u-v-(p_Y(u)-p_Y(v))=(u-p_Y(u))-(v-p_Y(v))$$ You can write $u-v=p_Y(u)-p_Y(v) + z$ and prove that $$\Vert u-v \Vert^2=\Vert p_Y(u)-p_Y(v) \Vert^2 + \Vert z \Vert^2 + 2 \mathcal{Re} \langle z, p_Y(u)-p_Y(v) \rangle $$ with $\mathcal{Re} \langle z, p_Y(u)-p_Y(v) \rangle  \ge 0$ considering the paragraph above.
Inequality (1) proves that the projection is continuous.
A: This is mostly an answer to the question in comments - "How do you prove that projection is linear?"
Claim (1): 

$T(Tx)=Tx$

Claim (2): 

$Tx$ is the unique element of $Y$ such that for all $y\in Y$, $\langle y,x-Tx\rangle = 0 $.

(1) is trivial. 
(2) requires more work. If $\langle y,x-Tx\rangle\neq 0$ for some $y\in Y$, then you can move in the direction $y$ from $Tx$ to get closer to $x$. I leave the details of that proof to you.
Showing uniquess: if $y_0\in Y$ has the above property then $x-Tx=x-y_0-(Tx-y_0)$ and:
$$\|x-Tx\|^2 = \|x-y_0\|^2 -2\langle x-y_0, Tx-y_0\rangle + \|Tx-y_0\|^2$$
But $\langle x-y_0, y_0-Tx\rangle=0$ by our assumptions, since $y_0-Tx\in Y$, which shows that $y_0$ is at least as close to $x$ as $Tx$ was. Since you've already shown that the definition of projection gives a unique element, this means $y_0=Tx$.

The characterization in (2) is what lets you show that $T$ is linear - that $T(x_1+x_2)=T(x_1)+T(x_2)$ and $T(\alpha x)=\alpha T(x)$.
Claim (3):

$T$ is linear.

Claim (4):

$\langle Tx,Tx\rangle = \langle Tx,x\rangle$

This follows since $\langle x-Tx,Tx\rangle=0$, from claim (2), since $Tx\in Y$.
