Can anyone help me with this proof.
Let $M=\{e_1, e_2, \ldots\}$ be an orthonormal subset of a Hilbert space $H$ and $A=\overline{\textrm{span}(M)}$. Show that the orthogonal projection $P:H\rightarrow A$ is given by, $$ Px=\sum_{i=1}^\infty \langle x, e_i\rangle e_i$$.
Suppose $x \in H$ and $u \in A$ are such that $$(x-u,v) = 0 \quad \forall v \in A.$$ Then for any $v \in A$ a quick calculation shows that $$\|u-x\|^2 - \|v - x\|^2 = 2(x-u,u-v) - \|u - v|^2 \le -\|u - v\|^2 \le 0$$ since $u-v \in A$. Thus $$\|u-x\|^2 \le \|v - x\|^2 \quad \forall v \in A$$ which shows that $u = Px$.
Now define $$u = \sum_{i=1}^\infty (x, e_i)e_i$$ Then for each $n$ the linearity and continuity of the inner product and the orthonormality of the $\{e_n\}$ yield $$(u-x, e_n) = (u,e_n) - (x,e_n) = \left(\sum_{i=1}^\infty (x,e_i)(e_i, e_n)\right) - (x, e_n)$$ $$ = (x,e_n)(e_n, e_n) - (x,e_n) = (x,e_n) - (x,e_n) = 0.$$ Invoking continuity and linearity of the inner product again, this shows that $$(u-x,v) = 0 \quad \forall v \in A$$ since $A = \overline{\mbox{span } \{e_n\}}$.
Combining the first two points shows that in fact $$Px = \sum_{i = 1}^\infty (x,e_i)e_i$$ as desired.