Let $X$ be the set of all the real- (or complex-) valued functions that are defined and continuous on a closed interval $[a, b]$ on the real line, where $a$ and $b$ are some fixed real numbers such that $a < b$, with the inner product defined by $$ \langle x, y \rangle := \int_a^b x(t) \overline{y(t)} dt. \tag{0} $$
Is this inner product space also a Hilbert space?
My Attempt:
Let $\left( x_n \right)_{n \in \mathbb{N} }$ be a Cauchy sequence in $X$. Then, given any real number $\varepsilon > 0$, there exists a natural number $N_\varepsilon$ such that $$ \left\lVert x_m, x_n \right\rVert = \sqrt{ \int_a^b \left\lvert x_m(t) - x_n(t) \right\rvert^2} < \varepsilon \tag{1} $$ for any natural numbers $m$ and $n$ such that $m > N_\varepsilon$ and $n > N_\varepsilon$. [Refer to (0) above.]
What next? How to proceed from here? Or, is there a Cauchy sequence that fails to converge in our inner product space $X$?
PS:
Based on the Hint given by @belkacemabderrahmane, here is what I do further:
For each $n \in \mathbb{N}$, let us define the function $x_n \colon [a, b] \longrightarrow \mathbb{C}$ by the formula $$ x_n(t) := \left( \frac{t-a}{b-a} \right)^n. \tag{2} $$ Then we find that, for any natural numbers $m$ and $n$, we have $$ \begin{align} & \ \ \ \left\lVert x_m - x_n \right\rVert \\ &= \sqrt{ \int_a^b \left\lvert \left( \frac{t-a}{b-a} \right)^m - \left( \frac{t-a}{b-a} \right)^n \right\rvert^2 dt } \\ &= \sqrt{ \int_a^b \left[ \left( \frac{t-a}{b-a} \right)^m - \left( \frac{t-a}{b-a} \right)^n \right]^2 dt } \\ &= \sqrt{ \int_a^b \left[ \left( \frac{t-a}{b-a} \right)^{2m} - 2 \left( \frac{t-a}{b-a} \right)^{m+n} + \left( \frac{t-a}{b-a} \right)^{2n} \right] dt } \\ &= \sqrt{ \frac{1}{(b-a)^{2m}} \int_a^b (t-a)^{2m} dt - \frac{2}{(b-a)^{m+n} } \int_a^b (t-a)^{m+n} dt + \frac{1}{(b-a)^{2n} } \int_a^b (t-a)^{2n} dt } \\ &= \sqrt{ \frac{1}{(b-a)^{2m}} \frac{(b-a)^{2m+1} }{ 2m+1} - \frac{2}{(b-a)^{m+n} } \frac{ (b-a)^{ m+n+1 } }{ m+n+1 } + \frac{1}{(b-a)^{2n} } \frac{ (b-a)^{2n+1} }{ 2n+1} } \\ &= \sqrt{ (b-a) \left[ \frac{1}{2m+1} - \frac{2}{m+n+1} + \frac{1}{2n+1} \right] }, \end{align} $$ which can be made as small as we please by taking $m$ and $n$ sufficiently large. Thus our sequence $\left( x_n \right)_{n \in \mathbb{N} }$ is indeed a Cauchy sequence.
Suppose that this sequnece converges to some point $x \in X$. Then we have $$ \lim_{n \to \infty} \sqrt{ \int_a^b \left\lvert x_n(t) - x(t) \right\rvert^2 } = 0, $$ which implies $$ \lim_{n \to \infty} \int_a^b \left\lvert x_n(t) - x(t) \right\rvert^2 = 0 $$ also.
How to proceed from here? How to show rigorously and in detail that this function $x$ is not in our space?
Let's take $[a,b] = [0,1]$ for simplicity. A better try would be to take the functions which are $$ x_n(t) = \begin{cases} 0 & t < 1/2 - 1/n\\ \text{linear} & 1/2-1/n \le t < 1/2 \\ 1 & 1/2 \le t. \end{cases} $$ This was a hint provided in another question that was linked in the comments. Suppose that $x$ is a continuous function such that $x_n\to x$ in $X$. Then for any $\epsilon > 0$, check for yourself $$ \int_0^{1/2-\epsilon}|x(t)|^2\,dt = 0, $$ and $$ \int_{1/2}^1|x(t)-1|^2\,dt = 0. $$ Since we assume $x$ is continuous (i.e., $x\in X$), by standard arguments, you can conclude that $x(t) = 1$ for $t\ge 1/2$, and $x(t) = 0$ for $t < 1/2$. Why is this a contradiction?