I assume that by compact, you mean quasi-compact (i.e. not necessarily Hausdorff but with the finite sucover property), otherwise any $Y$ with the coarsest topology would be a counterexample.
Then it seems that your notion is equivalent to pseudocompactness, namely that any continous $f:Y\to\mathbb{R}$ has compact image. I checked that quickly, so I may be wrong (I haven't found a reference in MR, although I would bet it is an exercise in Bourbaki).
The idea is first to observe that the property remains the same if you substitute $\mathbb{R}$ with $[0,1]\simeq[-\infty,\infty]$, basically because any one is included in the other (up to homeo).
Then since the graph of a continuous $f:Y\to\mathbb{R}$ is closed, $f(Y)$ must be closed in $\mathbb{R}$ and also $[-\infty,\infty]$, hence bounded. Similarly, any lower/upper semi-continuous $f$ is lower/upper bounded and attains its inf/sup.
For the converse, if $F$ is closed in $Y\times[0,1]$, but not its projection $F_2$, one may assume that $0$ isn't in $F_2$ but is in its closure. Then the function $y\mapsto \inf t : (y,t)\in F $ is lower semi-continuous and doesn't attain its infimum $0$.
By the way, pseudocompact doesn't imply (quasi-)compact, even for "nice" (completely regular) spaces, as is seen with the long line or simply the first uncountable ordinal $\omega_1$ (with order topology). There are also non-Hausdorff examples, see wikipedia article.