Example of discontinuous operator over an inner product space (If you want to skip the context, please go straight to the box describing the desired example. Thank you.)
I am taking a linear algebra course in a master's level, for which the textbook is mainly Hoffman and Kunze's book. Upon arriving at the definition of orthogonal complement of a set (in an inner  product space), the professor mentioned there might be one such operator. So I want to

Find an operator $ T:V \to V $ such that $T$ is linear, but discontinuous, with $V$ an
  inner product space (taken over the field of scalars
  $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$) with $\dim(V) = \infty$.

I have tried to come up with such an example, to no avail. He mentioned this might occur when $range(T)$ is not closed, and that we would need topological concepts for that discussion, and maybe for lack of time or interest, did not give such an example. I also could not imagine how the range of a linear operator taken over a field of real or complex scalars could not be closed.
We will most likely see that next semester in functional analysis, but I would like to see such an example, mainly because my linear algebra knowledge relies strongly on finite dimension spaces, so I am still building my intuition with infinite dimension spaces. My goal is to know more intuitively which of the results I already know for finite dimension spaces continue valid in infinite dimension, or to have a hint why they may not be valid.
 A: Consider the space of polynomials $V=\mathbb{C}[x]$ (you can also take reals as the base field) and then define $T=\delta$ where $\delta(f)$ is the formal derivative of $f$.
A: Take the inner product space $C[0, 1]$ (over $\Bbb{R}$) with expected inner product :
$$\langle f, g \rangle = \int_0^1 f(x) g(x) \, \mathrm{d}x.$$
Consider the valuation map $\phi$ at $x = 0$. Specifically,
$$\phi(f) = f(0).$$
This map is clearly linear, but it's not bounded. If we fix any $M > 0$, define
$$f : [0, 1] \to \Bbb{R} : f(x) = \begin{cases} M - M^3x & \text{if } 0 \le 1 \le \frac{1}{M^2} \\ 0 & \text{otherwise}.\end{cases}$$
Clearly $\phi(f) = M$, but
\begin{align*}
\langle M, M \rangle &= \int_0^{1/M^2} (M - M^3x)^2 \, \mathrm{d}x \\
&= \left[M^2x - M^4x^2 + \frac{1}{3}M^6x^3\right]_{x=0}^{1/M^2} \\
&= \frac{1}{3}.
\end{align*}
So, we can construct a function $f$, of a fixed distance $\frac{1}{\sqrt{3}}$ from $0$, so that $\phi(f)$ is larger than any given positive bound. Hence $\phi$ is unbounded, and hence discontinuous.
