It's natural to expect that there are linear operators on a pre-Hilbert space that are bounded but have no adjoint. What is an example?
Any bounded operator $T$ on a pre-Hilbert space $H$ has a unique extension to a bounded operator $\tilde{T}$ on the completion $\tilde{H}$ of $H$. Then $\tilde{T}$ has an adjoint, and if $T$ has an adjoint, i.e. there exists a bounded operator $T^\ast\colon H\to H$ with $\langle Tx,y\rangle_H = \langle x, T^\ast y\rangle_H$ for all $x,y\in H$, that must be the restriction of $(\tilde{T})^\ast$ to $H$, since we have $\langle Tx,y\rangle_H = \langle x, \tilde{T}^\ast y\rangle_{\tilde{H}}$.
So $T\colon H\to H$ has an adjoint if and only if $\tilde{T}^\ast(H) \subset H$.
That criterion makes it easy to construct an example. Let $H \subset \ell^2(\mathbb{N})$ the subspace of sequences with only finitely many nonzero terms. For any $\xi \in \ell^2(\mathbb{N})$, consider $T_\xi \colon x \mapsto \langle x,\xi\rangle_{\ell^2}\cdot e_1$. Since $$\langle T_\xi x, y\rangle_{\ell^2} = \langle \langle x,\xi\rangle_{\ell^2}\cdot e_1,y\rangle_{\ell^2} =\langle x,\xi\rangle_{\ell^2}\cdot \overline{y_1} = \langle x, y_1\cdot\xi\rangle_{\ell^2} = \langle x ,\langle y,e_1\rangle_{\ell^2}\cdot \xi\rangle_{\ell^2},$$
we have $\tilde{T}_\xi^\ast \colon y \mapsto \langle y,e_1\rangle_{\ell^2}\cdot \xi$, in particular $\tilde{T}_\xi^\ast(e_1) = \xi$. If $\xi \notin H$, the restriction of $T_\xi$ to $H$ is a bounded operator on $H$ without adjoint.
Generally, if $H$ is an incomplete pre-Hilbert space, many operators on $H$ will have the property $\tilde{T}^\ast(H) \not\subset H$, so working with adjoints on incomplete pre-Hilbert spaces doesn't look too fruitful (working with incomplete normed spaces generally is rarely done, one usually immediately considers their completion).
When does an unbounded operator on a Hilbert space have an adjoint?
When it is densely defined.
The characterisation of the adjoint by
$$\bigl(\forall x\in D(A)\bigr)\bigl(\forall y\in D(A^\ast)\bigr)\bigl(\langle Ax,y\rangle = \langle x, A^\ast y\rangle\bigr)\tag{1}$$
only determines $A^\ast y$ up to elements of $D(A)^\perp$, and since $(1)$ shall uniquely determine $A^\ast y$ for $y\in D(A^\ast)$, we need $D(A)^\perp = \{0\}$, i.e. $\overline{D(A)} = \{0\}^\perp = H$.
If for some $y\in H$ there exists a $z\in H$ such that for all $x\in D(A)$
$$\langle Ax,y\rangle = \langle x, z\rangle,\tag{2}$$
then $x \mapsto \langle Ax,y\rangle$ is a continuous linear form on $D(A)$ (with the topology induced by $H$).
Conversely, if $x\mapsto\langle Ax,y\rangle$ is continuous on $D(A)$, it extends to a continuous linear form on $H$, and by Riesz' theorem there is a $z\in H$ with $(2)$. That $z$ is uniquely determined by $(2)$ if and only if $D(A)$ is dense in $H$. In any case, the projection of such a $z$ in $\overline{D(A)}$ is uniquely determined by $(2)$, so viewing $A$ as an operator from a dense subspace of $\overline{D(A)}$ to $H$, its adjoint exists.
For an arbitrary densely defined operator, the adjoint is however often not useful. You can have
$$D(A^\ast) = \left\lbrace y \in H : (x\mapsto \langle Ax,y\rangle) \text{ is continuous}\right\rbrace = \{0\}.$$
The adjoint is more useful if it is itself densely defined. So when is $D(A^\ast)$ dense in $H$?
Before answering that, let's take a look how the graphs $\Gamma(A)$ of $A$ and $\Gamma(A^\ast)$ of $A^\ast$ are related. They are both linear subspaces of $H\times H$, which we endow with the inner product
$$\langle (x,y), (v,w)\rangle_{H\times H} =\langle x,v\rangle_H + \langle y,w\rangle_H$$
to obtain a Hilbert space structure. Then
$$\begin{align}
(y,z) \in \Gamma(A^\ast) &\iff \bigl(\forall x\in D(A)\bigr)\bigl(\langle Ax,y\rangle_H - \langle x, z\rangle_H = 0\bigr)\\
&\iff \bigl(\forall x\in D(A)\bigr)\bigl(\langle (x,-Ax),(z,y)\rangle_{H\times H} = 0\bigr)\\
&\iff (z,y) \in \Gamma(-A)^\perp.
\end{align}$$
So, with the coordinate swapping $S\colon H\times H \to H\times H; \; (x,y) \mapsto (y,x)$, we have
$$\Gamma(A^\ast) = S(\Gamma(-A)^\perp) = S(\Gamma(-A))^\perp.$$
If $A^\ast$ is densely defined, so $A^{\ast\ast}$ exists, then a short manipulation of the above shows that
$$\Gamma(A^{\ast\ast}) = \Gamma(A)^{\perp\perp} = \overline{\Gamma(A)},$$
so the closure of the graph of $A$ must be the graph of a linear operator - $A$ must be closeable.
Conversely, for any linear operator $B\colon D(B) \to H$, where $D(B) \subset H$, we can look at the subspace $R(B^\ast) := S(\Gamma(-B))^\perp$ of $H\times H$. This is the set of $(y,z) \in H\times H$ with
$$\bigl(\forall x \in D(B)\bigr)\bigl(\langle Bx, y\rangle_H = \langle x,z\rangle_H\bigr).$$
By the above, $R(B^\ast)$ is the graph of a linear operator (namely $B^\ast$) if and only if $B$ is densely defined.
So we have: Let $A\colon D(A) \to H$ be a densely defined linear operator. $A^\ast$ is densely defined if and only if $A$ is closeable, i.e. if and only if $\overline{\Gamma(A)}$ is the graph of a linear operator, then $$\Gamma(A^{\ast\ast}) = \overline{\Gamma(A)}.$$
Best Answer
Your argument does not prove the result at this point. For a bounded linear operator, $T$ is self-adjoint ($T^*=T$) iff $(Tx,y)=(x,Ty)$ for every $x,y\in H$. So you need to do more.
Using the textbook lemma is a good idea. It yields a straightforward argument. The proof of that lemma is not very difficult if you are familiar with polarization. It boils down to observing that $$ 2(Qx,y)=(Q(x+y),x+y)-i(Q(x+iy),x+iy)=0\qquad\forall x,y\in H. $$ Then for $y=Qx$, you get $(Qx,Qx)=\|Qx\|^2=0$, whence $Qx=0$ for every $x$, i.e. $Q=0$. So you just have to observe that $Q=T-T^*$ satisfies the condition $$(Qx,x)=(Tx,x)-(T^*x,x)=(Tx,x)-(x,Tx)=(Tx,x)-\overline{(x,Tx)}=(Tx,x)-(Tx,x)=0.$$ The lemma proves that $Q=T-T^*=0$, i.e. $T=T^*$ as desired.
You could also use polarization-like formulae to prove directly that $(Tx,y)=(x,Ty)$. But I think using this lemma is better.
Note that the lemma is false in the real case. Think of $\pi/2$ rotations. For instance, on $\mathbb{R}^2$: $$ Q=\pmatrix{0&-1\\1&0}\quad\Rightarrow\quad (Qx,x)=-x_2x_1+x_1x_2=0\quad\forall x\in\mathbb{R}^2. $$