These are not the usual definitions as I know them.$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$
First, I am only familiar with the situation that $H$ is a Hilbert space and $D(T)$ is dense in $H$ (which entails no loss of generality, as we can replace $H$ with the completion of $D(T)$.)
I would say:
$T$ is symmetric if $\inner{Tx}{y} = \inner{x}{Ty}$ for all $x,y \in D(T)$. (Note your definition doesn't make sense, because you are applying $T$ to vectors that may not be in $D(T)$.)
$T$ is Hermitian if it is symmetric and bounded. (If $T$ is bounded then it has a unique bounded extension to all of $H$, so we may as well assume $D(T) = H$ in this case.) Since a symmetric operator is always closable, the closed graph theorem implies that a symmetric operator with $D(T) = H$ is automatically bounded.
$T$ is self-adjoint if the following, more complicated condition holds. Let $D(T^*)$ be the set of all $y \in H$ such that $|\inner{Tx}{y}| \le C_y ||x||$ for all $x \in D(T)$, where $C_y$ is some constant depending on $y$. If $T$ is symmetric, one can show that $D(T) \subset D(T^*)$; $T$ is said to be self-adjoint if it is symmetric and $D(T) = D(T^*)$.
With these definitions, we have Hermitian implies self-adjoint implies symmetric, but all converse implications are false.
The definition of self-adjoint is rather subtle and this may not be the place for an extended discussion. However, I'd recommend a textbook such as Reed and Simon Vol. I. Perhaps I'll just say that symmetric operators, although the definition is simple, turn out not to be good for much, per se. One needs at least self-adjointness to prove useful theorems.
Let $\mathscr{H}$ be a Hilbert space. The domain of an operator $A$ on $\mathscr{H}$ is denoted $D(A)$; by an extension of $A$ is meant an operator $B$ with $D(A)\subset D(B)$ and $B|D(A)=A$ (where $B|D(A)$ denotes the restriction of $B$ to $D(A)$). If $B$ is an extension of $A$ it is very standard to write $A\subset B$.
Now take $A$ to be densely defined, meaning that the linear subspace $D(A)$ of $\mathscr{H}$ is dense in $\mathscr{H}$. This condition allows defining the adjoint operator $A^\ast$ of $A$; its definition is such that $D(A^\ast)$ is the set of all $y\in\mathscr{H}$ such that the map $D(A)\ni x\mapsto (Ax,y)\in\mathbf{C}$ is continuous (by a theorem of Hahn-Banach this map extends then to $\mathscr{H}$, and does so uniquely since we assumed $A$ to be densely defined). By the Riesz representation theorem for every $y\in D(A^\ast)$ there is a unique element of $\mathscr{H}$, which is denoted $A^\ast y$, such that $(Ax,y)=(x,A^\ast y)$ for all $x\in D(A)$. Thus $A^\ast$ is so defined as to guarantee $(Ax,y)=(x,A^\ast y)$ for all $x\in D(A)$ and $y\in D(A^\ast)$.
$A$ is symmetric (or formally self-adjoint, apparently physicists call them also Hermitian, but no mathematician would do this) if $A\subset A^\ast$; self-adjoint if $A=A^\ast$. Thus every self-adjoint operator is symmetric, but the converse need not hold. However, if $A$ is continuous and $D(A)=\mathscr{H}$ then $A$ symmetric implies $A$ self-adjoint. (In the finite-dimensional case every linear map is continuous.)
Something that is striking in quantum mechanics (theorem of Hellinger-Toeplitz): if $A$ is symmetric and $D(A)=\mathscr{H}$ then $A$ is continuous. Hence if $A$ is not continuous and symmetric, it cannot be defined on the whole of $\mathscr{H}$. (This shows that you cannot discuss quantum-mechanics without worrying about domains, since one can show that if operators $A$ and $B$ satisfy the canonical commutation relation $AB-BA=iI$ then at least one of $A$ and $B$ cannot be continuous).
Also a symmetric operator $A$ is self-adjoint iff its spectrum is a subset of the real line (this is important in quantum mechanics as the spectrum is there given a physical interpretation).
There is another notion which is frequently useful, particularly in mathematical physics. $A$ is essentially self-adjoint if $A$ is symmetric and its closure is self-adjoint (such an operator admits a unique self-adjoint extension, namely its closure). This appears e.g. in the standard representation of the canonical commutation relations: $\mathscr{H}=L^2(\mathbf{R})$, $A=-id/dx$ and $B$ be multiplication by $x$. $A$ and $B$ are essentially self-adjoint on the Schwartz space $\mathscr{S}(\mathbf{R})$.
Remark: Even if $A$ is densely defined, $A^\ast$ need not be densely defined (in fact it is densely defined iff $A$ is closable, e.g. if $A$ is symmetric).
If you want a reference, the standard reference is Reed/Simon: methods of modern mathematical physics, volume I (this is perfectly rigorous by the standards of mathematics).
Best Answer
Maybe this argument works?
Integrating by parts, we see that $$T^{\ast} \psi = i \frac{d}{dx}\psi.$$ Now let $j(x)$ be any positive, smooth function with suppose $(-1,1)$ such that $\int_{-\infty}^{\infty} j(x) =1$. Define $j_{\epsilon}(x) := \epsilon^{-1} j(x/\epsilon)$. Fix $0 < \alpha < \beta < 1$ and define \begin{eqnarray*} f_{\epsilon}^{\alpha, \beta}(x) &=& j_{\epsilon}(x - \beta) - j_{\epsilon}(x- \alpha), \\ g_{\epsilon}^{\alpha, \beta}(x) &=& \int_0^x f_{\epsilon}^{\alpha, \beta}(t)dt. \end{eqnarray*} Let $\psi \in D(T^{\ast})$. For $\epsilon$ sufficiently small, $g_{\epsilon}^{\alpha, \beta} \in D(T)$ and $$(Tg_{\epsilon}^{\alpha, \beta}, \psi ) = (g_{\epsilon}^{\alpha, \beta}, T^{\ast}\psi ).$$ As $\epsilon \to 0$, $g_{\epsilon}^{\alpha, \beta} \to \chi_{(\alpha, \beta)} \in L^2(0,1)$, so $$(g_{\epsilon}^{\alpha, \beta}, T^{\ast} \psi) \longrightarrow - \int_{\alpha}^{\beta} (T^{\ast}\psi)(x) dx.$$ Moreover, if $\varphi \in C[0,1]$, then $$J_{\epsilon}\varphi = \int_0^1 j_{\epsilon}(x-t) \varphi(t)dt \xrightarrow{ \ \ L^2 \ \ } \varphi.$$ Further, we claim that $\| J_{\epsilon} \| \leq 1$. To see this, suppose that $\psi \in L^2[0,1]$, then \begin{eqnarray*} \left| (\psi, J_{\epsilon}\varphi) \right| & \leq & \iint j_{\epsilon}(x-t) \left| \varphi (t) \right| \left| \psi (x) \right| dx dt \\ &=& \iint j_{\epsilon}(y) \left| \varphi (t) \right| \left| \psi(y+t) \right| dy dt \\ & \leq & \| \varphi \| \| \psi \| \int j_{\epsilon}(y) dy \\ &=& \| \varphi \| \| \psi \|. \end{eqnarray*} By a simple $\epsilon/3$ argument, $J_{\epsilon} \varphi \to \varphi$ in $L^2$ for all $\varphi \in L^2[0,1]$. So $(Tg_{\epsilon}^{\alpha, \beta}, \psi) \to -i(\psi(\beta)- \psi(\alpha))$ in mean square as $\epsilon \to 0$. Therefore, for almost every $\alpha, \beta$, $$i(\psi(\beta) - \psi(\alpha)) = \int_{\alpha}^{\beta} (T^{\ast}\psi)(x)dx.$$ So $\psi$ is absolutely continuous and $$i \frac{d}{dx}\psi(x)= (T^{\ast}\psi)(x).$$ Conversely, integrating by parts, we see that if $\psi \in AC[0,1]$, then $\psi \in D(T^{\ast})$ and $T^{\ast} \psi = i \frac{d}{dx}$. We therefore conclude that $T^{\ast} = i d/dx$ and $D(T^{\ast}) = AC[0,1]$.
Reference: R&S 1.