First, note that the definition of almost continuous is very similar to the definition of continuous, only slightly weaker - namely, a function $\,f:X\to Y$ is continuous at $x\in X$ iff for any open set $V\subset Y$ containing $\,f(x)$, we have the $\,f^{-1}(V)$ (not just its closure) is a neighbourhood of $x$. It should be immediately obvious that if $\,f$ is continuous at $x$, then $\,f$ is almost continuous at $x$. Applying this to your question, you can already see that $\,f$ is almost continuous, at all but countably many points (namely, the discontinuities).
Now, suppose $x$ is one of the discontinuities of $\,f$. Consider any open set $V\subset Y$ containing $\,f(x)$. Now, $\,f^{-1}(V)$ is not necessarily a neighbourhood of $x$, but what can we say about the closure of $\,f^{-1}(V)$?
Let $S=\{1/n,n\in \mathcal N\}$. Now consider, $\,f:[0,1]\to \{0,1\}$ given by the indicator function on $S$, i.e., $\,f(x)=1$ if $x \in S$, $0$ otherwise. Clearly, $\,f$ has countably many discontinuities, namely at all the points of $S$ and no others. Now, suppose we wish to consider an open set $V\subset \{0,1\}$ containing $\,f(1/2)=1$, say the open set $\{1\}$. Now, $\,f^{-1}(\{1\})=S$. Thus, the closure of $\,f^{-1}(\{1\})$, i.e., the closure of $S$ is $S\cup \{0\}$, which is not a neighbourhood of $1/2$. Therefore, $f$ is not almost continuous at $1/2$ (or in fact, by similar arguments at any of the discontinuities in $S$) and so your question cannot be answered in the positive.
