I think the first FTC:
If $f: [a,b] \to \Bbb R$ is continuous then $F: [a,b] \to \Bbb R$ defined by $F(x)=\int_a^x f(t)dt$ is differentiable and $F'(x)=f(x)$ for all $x \in [a,b]$.
is what people mean by saying the integration (which defines $F$) is the inverse of differentiation (as we have found a function with derivative $f$).
The second FTC
If $f: [a,b] \to \Bbb R$ is Riemann-integrable on $[a,b]$ and we have a function $F: [a,b] \to \Bbb R$ such that $F'(x)=f(x)$ on $[a,b]$, then $\int_a^b f(x)dx=F(b)-F(a)$.
is more of a "recipe" to find an integral: the target is to compute the definite integral and the tool we're given is to find an antiderivative. So not an inverse as such but a method. It's a bit of an iffy one, as an antiderivative $F$ need not exist at all (except when $f$ is continuous and the first FTC gives us one, but not explicitly, but at least we know some solution exists, but we don't have it in computable form yet). I think the first is closer to giving a direct "inverse" connection between integration and differentiation (and is often used in other contexts when we differentiate wrt boundaries of integrals, etc.). But that's just one view.
The first FTC can be summarised as $$\frac{d}{dx}\int_a^x f(t)dt = f(x)$$ so "Applying the integration operator to $f$, followed by the differentiation operator gives us back $f$ again".
What you have written is true, but it is not the fundamental theorem of calculus. You've just rearranged the definition
$$
F(x) = \int_c^x f(t) dt
$$
using integral properties.
The First Fundamental Theorem of Calculus
If we define $F(x) = \int_c^x f(t) dt$, then one result worthy of the name "fundamental theorem of calculus" says
$$
F'(c) = f(c)
$$
whenever $f$ is continuous at $c$. Note that we need $f$ to be continuous at $c$ or else the result can fail to be true.
Why is this true? You've made the key insight.
$$
F'(c) = \lim_{h \to 0} \frac{F(c + h) - F(c)}{h} = \lim_{h \to 0} \frac{\int_c^{c + h} f(t) dt}{h}
$$
To finish the proof, you just need to explain why that last limit is equal to $f(c)$. This is where you will have to use continuity at $c$.
The Second Fundamental Theorem of Calculus
The other result that goes by the name "fundamental theorem of calculus" says that if we have a function $F$ such that $F' = f$ and $f$ is integrable on $[a,b]$, then
$$
\int_a^b f(t) dt = F(b) - F(a)
$$
The classic proof uses a completely different trick. We use the mean value theorem with $F$ and $f$ and the integrability of $f$ to set up a telescoping sum.
People will often give a simpler proof which is more properly a corollary of the first fundamental theorem. If we assume $f$ is continuous on $[a,b]$, then we can argue as in RafGaming's answer to prove the same result. But note that the mean value theorem proof, though more complicated, proves the result without making any assumptions about the continuity of $f$.
Comparing the Two Theorems
Suppose $f$ is integrable on $[a,b]$ and $c \in (a,b)$
Schematically, in the first theorem, we have
$$
\left[ F(x) = \int_c^x f(t) dt \quad \& \quad \lim_{x \to c} f(x) = f(c) \right] \implies F'(c) = f(c)
$$
Note, again, that we need continuity of $f$ at $c$. The statement $F(x) = \int_c^x f(t) dt \implies F'(c) = f(c)$ is not true.
In the second theorem, we have
$$
F' = f \quad \implies \int_a^b f(t) dt = F(b) - F(a)
$$
Though similar looking, these results are not the same. In fact, they are not even exactly converses. They're two distinct results that both formalize the intuition that "differentiation undoes integration and vice versa."
Best Answer
I would suggest you to have a look at this answer where I have discussed the Fundamental Theorem of Calculus for functions which are not necessarily continuous.
Next note that the integrand here has a discontinuity at $x=1$ and as far as the interval of integration is concerned the discontinuity is removable and hence we just remove it by changing value of integrand at $x=1$ to $0$. Doing this has no impact on the value of the integral (because the value of a Riemann integral does not depend on the value of the integrand at a finite number of points in the interval under consideration) and thus the desired integral is equal to $\int_{1/2}^{1}0\,dx=0$.
In general if we have to evaluate an integral of the form $\int_{a} ^{b} f(x) \, dx$ where $f$ has a finite number of discontinuities at $x_{1},x_{2},\dots,x_{n}$ in $[a, b] $ with $$a\leq x_{1}\leq x_{2}\leq \dots\leq x_{n} \leq b$$ such that left and right hand limits of $f$ exist at each point of discontinuity then we split the integral as follows $$\int_{a} ^{b} f(x) \, dx=\int_{a} ^{x_{1}}f(x)\,dx+\int_{x_{1}}^{x_{2}}f(x)\,dx+\cdots+\int_{x_{n-1}}^{x_{n}}f(x)\,dx+\int_{x_{n}}^{b}f(x)\,dx$$ Each integral on right can be evaluated by observing that the integrand has removable discontinuity at the end points $x_{i}$ and the integrand can be redefined at these points to make it continuous on that interval.
You can use this technique to evaluate $\int_{1}^{2}[x^{2}]\,dx$ as $$\int_{1}^{2}[x^{2}]\,dx=\int_{1}^{\sqrt{2}}1\,dx+\int_{\sqrt{2}}^{\sqrt{3}}2\,dx+\int_{\sqrt{3}}^{2}3\,dx$$
Also to answer your question about evaluating the integral under consideration via second part of fundamental theorem of calculus, note that there is no anti-derivative of $[x] $ on interval $[1/2,1]$ (why? perhaps you should answer this yourself, but let me know if you feel issue here) and hence we can't use fundamental theorem of calculus here.
Also note that continuous functions are guaranteed to have an anti-derivative, but continuity is not necessary. There are discontinuous functions which possess anti-derivative and if we have a function $f$ which is Riemann integrable on $[a, b] $ and possesses an anti-derivative $F$ (note that existence of anti-derivative is not guaranteed and hence we are assuming its existence here as a part of the hypotheses) such that $F'(x) =f(x) $ for all $x\in[a, b] $ then we have $\int_{a}^{b}f(x)\,dx=F(b)-F(a)$.