Here's a slightly old-fashioned way of looking at topology. Let $X$ be a set. For each $x$ in $X$, we have a non empty set $N(x)$ of subsets of $X$, which we think of as being the set of neighbourhoods of $x$. The set $N(x)$ is required to satisfy some intuitive properties:
If $U$ is a neighbourhood of $x$, then $x \in U$.
The whole set $X$ is a neighbourhood of $x$.
If $U$ is a neighbourhood of $x$, and $U \subseteq V$, then $V$ is a neighbourhood of $x$.
If $U$ and $V$ are neighbourhoods of $x$, so is $U \cap V$.
If $U$ is a neighbourhood of $x$, then there is a subset $V$ such that $U$ is a neighbourhood of every point in $V$.
In more sophisticated language, we are saying that $N(x)$ is a filter of the set of subsets of $X$ containing $x$, satisfying the additional axiom (5). Note that there usually isn't a smallest neighbourhood: for example, you should agree that each interval $(x - \epsilon, x + \epsilon)$ is a neighbourhood of $x$, but the only set contained in all neighbourhoods of $x$ is $\{ x \}$, which can't possibly be a neighbourhood of $x$ as it contains no neighours of $x$!
Now, let $Y$ be any subset of $X$. The interior of $Y$ is defined to be the subset $Y^\circ$ of all $y$ in $Y$ such that $Y$ contains a neighbourhood of $y$. This seems reasonable enough: the interior of the closed interval $[a, b]$ is $(a, b)$, because the closed interval $[a, b]$ does not contain enough neighbours of $a$ or of $b$. We say that a subset $U$ is open in $X$ if $U^\circ = U$. The set of all open subsets of $X$ has the following pleasant properties:
The whole set $X$ is open, and the empty set is open.
If $\{ U_\alpha : \alpha \in I \}$ is a set of open subsets of $X$, then the union $\bigcup_\alpha U_\alpha$ is also an open subset of $X$.
If $U$ and $V$ are open, then $U \cap V$ is also open.
Exercise. Derive these properties of open sets from the neighbourhood axioms.
If you've ever seen a modern definition of topological space, you'll recognise these properties. It turns out that knowing the set of all open subsets gives you exactly the same information as knowing $N(x)$ for every point $x$ in $X$. This leads to somewhat less intuitive developments, like topological spaces without points, but let's leave that aside for now.
What else can we do with neighbourhoods? Well, we can talk about forms of convergence. For example, let $x_1, x_2, x_3, \ldots$ be a sequence of points in $X$, and let $x$ be a point. We say that $(x_n)$ converges to $x$ just if for any neighbourhood $U$ of $x$, there is a positive integer $N$ such that $x_n \in U$ for all $n > N$. Less formally, we are saying that a sequence converges to $x$ if it is eventually always in a neighbourhood $U$ of $x$, no matter how small we pick $U$ to be! More generally, if $Y$ is any subset of $X$, a point $x$ is a limit point of $Y$ if every neighbourhood of $x$ has a non-empty intersection with $Y$. The set $\overline{Y}$ of all limit points of $Y$ is called the closure of $Y$, and a set $Y$ is closed just if $\overline{Y} = Y$.
Exercise. Show that $Y$ is closed if and only if its complement $X \setminus Y$ is open.
And, of course, we can always talk about continuity. Let $X$ and $X'$ be topological spaces. A continuous map is a map $f : X \to X'$ with the following property: if $x$ is a limit point of a subset $Y$ of $X$, then $f(x)$ is a limit point of the image of $Y$ in $X'$. This captures the intuition that a continuous map should map nearby points to nearby points: after all, if $x$ is a limit point of $Y$, then $x$ is near $Y$ in some sense, so $f(x)$ should be near $f(Y)$. In other words, the image of the closure should be contained in the closure of the image, i.e. $f(\overline{Y}) \subseteq \overline{f(Y)}$.
Exercise. Show that $f : X \to X'$ is continuous if and only if $f^{-1} U'$ is open in $X$ for every open subset $U'$ of $X'$.
Hopefully these definitions are more convincing than the usual ones. If they are, then great – because they are exactly equivalent to the usual definitions!