This problem is best solved by drawing diagram by plotting the line $x+y=z$ in 'x-y' plane and measuring the area under the line over the rectangle enclosed by $(0,0), (0,1), (1,0)$ and $(1,1)$ [covering $(x+y\leqslant z)$]. This is my recommendation. Alternatively, you may want to use method of convolution while calculating cdf of the sum X+Y exactly the way you have attempted (for details see this).
Looking at your approach I can see the issue in the range of y used in the integral when you try to find $P(X+Y\le a)$ for $0<a<1$. Unfortunately you can't write: $\int_{0}^{1} \left( \int_{0}^{a-y} f(x) \ \mathrm{d} x \right) f(y) \ \mathrm{d}y = \int_{0}^{1} \left( a-y \right )1 \ \mathrm{d}y$, since, when $0<X<a-y$, $0<a-Y<1$; thus, $a-1<Y<a$; but since we are evaluating the cdf for $0\le a\le1$, the range of Y eventually reduces to $0<Y<a$.
Similar sort of issue appears for the case $1\le a\le2$. Basically we just need to be a bit careful about the ranges so that the bounds used are valid ones.
Suppose that the random variables are iid uniform over the interval $[0,1]$. The pdf is a flat line because each of the $X_{i}$ are equally likely to be anywhere between $[0,1]$. However, if you collect $n$ of them and look at the minimum, it is more likely to, for example, be towards the lower half of the interval than the upper half so it's pdf should have more area towards $0$ and less up by $1$. i.e. It won't be a flat line anymore!
We can represent this problem using the following experiment:
Draw a number line and put $y$ on it. $y$ can be any value but thinking about the range of the function $min(X_{1},...,X_{n})$, $y$ should sensibly be a positive real number in the interval [0,1]. Now imagine the different number of ways you can drop $n$ data points around $y$ so that the minimum value of all points (when considered together) is below $y$. There are a lot of ways to do this but you must keep in mind that each point must be located between the interval [0,1] since the probability of them appearing elsewhere outside of the closed interval [0,1] is 0.
Your instructor is giving you a good hint. It is almost always easiest to use cdfs when trying to find the distribution of mins and maxes. Note that
$$
F_{Y(y)} = P(Y \leq y) = P(\min (X_{1}, X_{2}, \ldots, X_{n}) \leq y).
$$
The idea is to relate this to probabilities involving the individual $X_{i}$ since you know things about them.
Now consider writing:
$$
P(\min (X_{1}, X_{2}, \ldots, X_{n}) \leq y) = 1 - P(\min (X_{1}, X_{2}, \ldots, X_{n}) > y).
$$
The only way to arrange your points so that $\min (X_{1}, X_{2}, \ldots, X_{n}) > y$ is to put them all above $y$.
So,
$$
\begin{array}{lcl}
F_{Y(y)} &=& P(Y \leq y) = P(\min (X_{1}, X_{2}, \ldots, X_{n}) \leq y)\\
&=& 1-P(\min (X_{1}, X_{2}, \ldots, X_{n}) > y)\\
&=& 1-P(X_{1}>y, X_{2}>y, \ldots, X_{n}>y)
\end{array}
$$
Now use the fact that the $X_{i}$ are independent to break that probability up into a product. Use the fact that they are identically distributed in order to write it as one probability to the $n$th power.
You are almost there since you can relate that one probability to the cdf of your original distribution!
Best Answer
We compute the density of $Z$ by convolution. Let $f_{X_1}(t)=\mathsf 1_{(0,1)}(t)$ and $f_{X_2}(t)=\mathsf 1_{(0,1)}(t)$ be the densities of $X_1$ and $X_2$, respectively. Then the density of $Z=X_1+X_2$ is given by \begin{align} f_Z(t) &= (f_{X_1}\star f_{X_2})(t)\\ &= \int_{\mathbb R}f_{X_1}(\tau)f_{X_2}(t-\tau)\ \mathsf d\tau.\\ &= \int_{\mathbb R}\mathsf 1_{(0,1)}(\tau)\mathsf 1_{(0,1)}(t-\tau)\ \mathsf d\tau. \end{align} For the integrand to be positive, we must have $0<\tau<1$ and $0<t-\tau<1$. First note that because $X_1$ and $X_2$ take values in $(0,1)$, $Z$ takes values in $(0,2)$. So for $0<t<1$ we have $0<\tau<t$, so that $$ f_Z(t) = \int_0^t \ \mathsf d\tau = t. $$ For $1<t<2$ we have $t-1<\tau<1$, so that $$ f_Z(t) = \int_{t-1}^1 \ \mathsf d\tau = 2 - t. $$ It follows that $$ f_Z(t) = t\cdot\mathsf 1_{(0,1)}(t) + (2-t)\cdot\mathsf 1_{(1,2)}(t). $$ We compute the distribution function of $Z$ by integrating the density. For $0\leqslant t<1$ we have $$ F_Z(t) = \int_0^t s\ \mathsf ds = \frac12 t^2. $$ For $1\leqslant t<2$ we have \begin{align} F_Z(t) &= \int_0^1 s\ \mathsf ds + \int_1^t (2-s)\ \mathsf ds\\ &= \frac12 + \frac12(t-1)(3-t)\\ &= \frac12 (1 +(t-1)(3-t)). \end{align} It follows that $$ F_Z(t) = \begin{cases} 0,&t<0\\ \frac12t^2,&0\leqslant t<1\\ \frac12 (1 +(t-1)(3-t)),&1\leqslant t<2\\ 1,&t\geqslant 2. \end{cases} $$